This is probably common knowledge, alas I have to confess my ignorance.


In simpler more abstract language, does $\mathcal{O}\_K$ being simply connected (having trivial etale $\pi\_1$) imply $\mathcal{O}\_K=\mathbb{Z}$?



Is there a ring of integers except for Z, such that every extension of it is ramified?
Yes, there are examples and Minkowski's proof for ${\mathbf Q}$ can be adapted to find a few of them. Some examples of this kind among quadratic fields $F$, listed in increasing size of discriminant (in absolute value), are 
$$
{\mathbf Q}(\sqrt{-3}), \ \ {\mathbf Q}(i), \ \ {\mathbf Q}(\sqrt{5}), \ \ {\mathbf Q}(\sqrt{-7}), \ \ {\mathbf Q}(\sqrt{2}), \ \ {\mathbf Q}(\sqrt{-2}).
$$
A cubic and quartic field $F$ that will come out of the method I describe below are ${\mathbf Q}(\alpha)$ where $\alpha^3 - \alpha - 1 = 0$ and ${\mathbf Q}(\zeta\_5)$. 


Now for the details. I suggest when reading this through for the first time that you keep a concrete example in mind, like $F = {\mathbf Q}(i)$. (That's what I did the first time I worked this out.) 


Over the rationals, Minkowski showed a number field with degree larger than 1 must have a discriminant whose absolute value is larger than 1. Over other number fields $F$ besides the rationals, the goal is to find sufficient conditions on $F$ so that *any* finite extension $E/F$ with $[E:F] > 1$ has its discriminant ideal ${\mathfrak d}\_{E/F}$ not equal to the unit ideal, and then a prime ideal factor will ramify in $E$. 


Rather than show ${\mathfrak d}\_{E/F} \not= (1)$, we will look for a sufficient condition on $F$ which assures us that the norm of this ideal is not 1. That means absolutely the same thing, but it's easier to work with ideal norms since they are positive integers rather than ideals, and moreover it lets us express the problem in terms of discriminants of number fields: the discriminants of $E$ and $F$ are related by $$|d\_E| = {\rm N}({\mathfrak d}\_{E/F})|d\_F|^{[E:F]}.$$ 
So aiming to show ${\mathfrak d}\_{E/F} \not= (1)$ is the same as *avoiding* 
$|d\_E| = |d\_F|^{[E:F]}$, which is the same as *avoiding* $$|d\_E|^{1/[E:{\mathbf Q}]} = |d\_F|^{1/[F:{\mathbf Q}]}.$$ We want sufficient conditions on $F$ to guarantee this equation for any proper finite extension $E/F$ can't take place. 


For any number field $K$, the quantity $|d\_K|^{1/[K:{\mathbf Q}]}$ is called the root discriminant of $K$. When $n = [K:{\mathbf Q}]$, Minkowski's lower bound on $|d\_K|$ is 
$$
|d\_K| \geq \left(\frac{\pi}{4}\right)^n\frac{n^{2n}}{n!^2},
$$
so we get the root discriminant lower bound
$$
|d\_K|^{1/n} \geq \left(\frac{\pi}{4}\right)\frac{n^{2}}{n!^{2/n}}.
$$
Call the right side $f(n)$, so the Minkowski bound says $|d\_K|^{1/[K:{\mathbf Q}]} \geq f([K:{\mathbf Q}])$. 


For $n = 1, 2, 3, 4$ the values of $f(n)$ are $.785, 1.570, 2.140, 2.565$, so we guess $f(n)$ is increasing and it's left as an exercise to prove that. (Hint: use the one-sided Stirling estimate $n! > (n/e)^n\sqrt{2\pi{n}}$.)


Returning to the extension $E/F$, let $m = [F:{\mathbf Q}]$, so $[E:{\mathbf Q}] = [E:F][F:{\mathbf Q}] \geq 2m$ since $E$ is a larger field than $F$. Now *if* 
$E/F$ had trivial discriminant ideal, we would have 
$$
|d\_F|^{1/[F:{\mathbf Q}]} = |d\_E|^{1/[E:{\mathbf Q}]} \geq f([E:{\mathbf Q}]) \geq f(2m).
$$
This hypothetical lower bound on the root discriminant of $F$ is larger than the proved Minkowski bound of $f(m)$. Suppose $F$ is a number field of degree $m$ 
whose root discriminant is less than $f(2m)$. If $E/F$ is unramified at all primes in $F$ 
then $E$ and $F$ have equal root discriminants, so the root discriminant of $E$ is less than $f(2m)$. However, we saw above that the root discriminant of $E$ is $\geq f(2m)$ when $[E:F] \geq 2$, so the only choice is $E = F$, i.e., no proper finite extension of $F$ can be unramified at all primes in $F$.


Our goal now is to find examples of number fields $F$ with degree $m$ whose 
root discriminant is smaller than $f(2m)$: $|d\_F|^{1/m} < f(2m)$. This is the same as 
$$
|d\_F| < f(2m)^m = \frac{\pi^mm^{2m}}{(2m)!}.
$$
Any $F$ which fits this condition will be an example. 


As a reality check, let $F$ be the rationals, so $m = 1$. We have 
$f(2) = \pi/2$ and $|d\_{\mathbf Q}| = 1 < \pi/2$, so $\mathbf Q$ has no unramified extensions. We're on the right track. 


Taking $m = 2$ we want $|d\_F| < f(4)^2 = 6.57$ and the fields ${\mathbf Q}(i)$, ${\mathbf Q}(\sqrt{-3})$, and ${\mathbf Q}(\sqrt{5})$ all work. 


If the root discriminant of $F$ is not below $f(2m)$, where $m = [F:{\mathbf Q}]$, 
we can still squeeze out some information, namely an upper bound on the degree of an everywhere (= at finite places) unramified extension of $F$. For instance, ${\mathbf Q}(\sqrt{2})$ has discriminant 8, which is not below 6.57, so this argument doesn't show on its own that every proper extension of ${\mathbf Q}(\sqrt{2})$ is ramified at some prime in ${\mathbf Q}(\sqrt{2})$. However, we can bound the degree of such an extension. The root disc. of ${\mathbf Q}(\sqrt{2})$ is $\sqrt{8} \approx 2.828$, which lies between $f(4)$ and $f(5)$, so any proper unramified extension of ${\mathbf Q}(\sqrt{2})$ must be a quadratic extension of ${\mathbf Q}(\sqrt{2})$. Quadratic extensions are automatically abelian, so we would have an abelian extension of ${\mathbf Q}(\sqrt{2})$ unramified at no finite places, and there's no such thing by class field theory since ${\mathbf Q}(\sqrt{2})$ has wide class number 1 (the class number not involving ramification at infinity). Therefore you can add ${\mathbf Q}(\sqrt{2})$ to the list of number fields with no proper extension unramified at all finite places. The same arguments apply to ${\mathbf Q}(\sqrt{-2})$ and ${\mathbf Q}(\sqrt{-7})$, whose root discriminants are also between $f(4)$ and $f(5)$.


Unfortunately, this method is really limited because although $f(n)$ is increasing, 
it's actually bounded. By Stirling's formula, $f(n)$ has limit $\pi e^2/4 \approx 5.803$. So when a number field $F$ has root discriminant exceeding this value, this method won't give us any examples at all (you can't find an $m$ for which the root discriminant is below $f(2m)$). [Edit: Using Odlyzko-type lower bounds on discriminants, one can produce some more examples as Torsten points out.]


This kind of argument using Minkowski's bound does work for a few cubic fields and quartic fields: for cubic fields it works as long as the discriminant of the field (in absolute value) is less than 31.39, and there are two such fields: ${\mathbf Q}(\alpha)$ and ${\mathbf Q}(\beta)$ where $\alpha^3 - \alpha - 1 = 0$ (discriminant -23) and $\beta^3 + \beta + 1 = 0$ (discriminant -31). The next smallest absolute value of a discriminant of a cubic field is 44, which is above the bound. For quartic fields we need the discriminant to be less than 158.32, and I know of three fields which work: ${\mathbf Q}(\gamma)$ where 
$\gamma^4 + 2\gamma^3 + 3\gamma + 1 = 0$ (discriminant 117), ${\mathbf Q}(\zeta\_5)$ has discriminant 125, and ${\mathbf Q}(\zeta\_{12})$ has discriminant 144. 


