Let $R$ be an integral domain such that every non-constant, irreducible polynomial $f(X) \in R[X]$ has degree $1$. 



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> **Q.** is it true that $R$ is a field?
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If $0 \ne a \in R$ , then $X^2-a$ is reducible in $R[X]$. Since this polynomial has content $1$ , we must have a factorization into one degree polynomials $X^2-a=(cX+d)(eX+g)=ecX^2+(de+cg)X+dg$. So $ec=1, de+cg=0, dg=-a$. So $d=-c^2g$, hence $a=-dg=(cg)^2$ . So every element of $R$ is a perfect square in $R$ . So if $R$ is a factorization domain with only irreducible non-constant polynomials in $R[X]$ being of degree $1$, then $R$ has no irreducible elements , hence $R$ must be a field. 


I have no idea about what happens if $R$ is not a factorization domain. 



Integral domain over which any non-constant, one variable, irreducible polynomial has degree 1
The answer is *no*, as shown by the following example.


Take as $R$ the ring $\bar{\mathbb{Z}}$ of [algebraic integers](https://en.wikipedia.org/wiki/Algebraic_integer). Since its fraction field is $\bar{\mathbb{Q}}$, which is algebraically closed, it follows by [Gauss lemma for GCD domains](https://en.wikipedia.org/wiki/Gauss%27s_lemma_(polynomial)) that the only non-constant, irreducible polynomials in $\bar{\mathbb{Z}}[x]$ have degree $1$. 


**Remark.** The ring $\bar{\mathbb{Z}}$ is non-Noetherian, for instance because there is an infinite ascending chain of (principal) ideals $$(2) \subset (2^{1/2}) \subset (2^{1/4}) \subset \ldots $$


Since it is a [Bézout domain](https://en.wikipedia.org/wiki/B%C3%A9zout_domain), it follows that it is not a UFD; in fact, for a Bézout domain, being Noetherian, PID and UFD are three equivalent conditions. 


