What is the set $\{x\in\Bbb R\mid \forall q\in\Bbb Q: q^x\in\Bbb Q\}$? What is the set $\{x\in\Bbb R\mid \forall q\in\Bbb Q: q^x\in\Bbb Q\}$?
Of course $\Bbb Z$ is a subset of this set.
Are there any other? if not what is the proof? is there a good reference for it?
 A: It is known that if $2^c$, $3^c$, and $5^c$ are all rational then $c$ is a nonnegative integer. It follows from the Six Exponentials Theorem, q.v. 
On the 1971 Putnam, it was asked to show that if $2^c,3^c,4^c,\dots$ are all integers then $c$ is a nonnegative integer. 
A: Expansion of Gerry Myerson/MJD answer.  
If $x$ is rational but not integer, then $2^x$ is irrational by unique factorization.  Now suppose $x$ is irrational.
Let's use the "Six Exponentials Theorem", which states:
Let $x_1, x_2, x_3, y_1, y_2$ be complex numbers, assume $x_1, x_2, x_3$ are linearly independent over $\mathbb Q$ and $y_1, y_2$ are linearly independent
over $\mathbb Q$.  Then at least one of $\exp(x_iy_j)$ is transcendental.  
Take $x_1 = \log 2, x_2 = \log 3, x_3 = \log 5, y_1 = 1, y_2 = x$.  Then $\log 2, \log 3, \log 5$ are linearly independent by unique factorization.  And $1,x$ are linearly independent since $x$ is irrational.  Conclusion: at least one of
$$
2, 3, 5, 2^x, 3^x, 5^x
$$
is transcendental.  So, in particular, they are not all rational.
A: Note that $-1$ is rational.  If $x$ satisfies the condition, in particular $(-1)^x$ is rational.  Taking that to mean
$e^{i\pi x}$ is rational, and thus real, I conclude that $x$ is an integer.
