It's straightforward that $t$ must be irrational. I have googled many variations of this question and browsed through some books on transcendental number theory. There is much that is said about when the base is the same, but not for when the power is the same like here.


For example, does $-\sqrt 2$ satisfy this? I would be happy even with some buzz words about what machinery covers this question, or whether it is part of some grand and long-term conjecture in transcendental number theory now far beyond our powers. 



For what real $t$ is $\{n^t : n \geq 1\}$ linearly independent over $\mathbb{Q}$?
Robert Israel's guess that this is true for $t$ an algebraic irrational does actually follow from Schanuel's conjecture.


Apply Schanuel's conjecture to $\log p$ and $t \log p$ for $n$ different primes $p$. These are linearly independent unless $t$ is the ratio of the logarithms of two integers, which it isn't because $t$ is an algebraic irrational.


So we get the transcendence degree of the following field is $2n$:


$$\mathbb Q( \log p\_1,\dots \log p\_n, t \log p\_1, \dots, t \log p\_n, p\_1, \dots, p\_n, p\_1^t, \dots, p\_n^t)$$


Now $p\_1,\dots, p\_n$ are in $\mathbb Q$ and $t \log p\_1, \dots , t \log p\_n$ are algebraic over, $\log p\_1, \dots, \log p\_n$, so the transcendence degree of this field is also $2n$:


$$\mathbb Q( \log p\_1,\dots \log p\_n, p\_1^t, \dots, p\_n^t)$$


Hence all these numbers are algebraically independent. If primes raised to the power $t$ are algebraically independent, it implies that natural numbers to the power $t$ are linearly independent over $\mathbb Q$ by factoring natural numbers into primes.


This is a special case of the general principle that everything follows from Schanuel's conjecture.


