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.

finiteset. $\endgroup$ – Emil Jeřábek Apr 28 '15 at 10:56