Title says it all. It's simple to find irrational numbers $x$ and $y$ such that $x^y$ is rational. If we want to have both $x^y$ and $y^x$ be rational, we can make the constraint 'easier' by just requiring $x=y$ so that the two conditions are equivalent, and then solving (for instance) the equation $x^x = 2$. But what if we have two distinct numbers?
My intuition says no, of course -- that there are irrational $x \neq y$ with $x^y, y^x \in \mathbb{Q}$. But I'm not sure how to show this. My first thought is taking something like $y=2x$, so that $$x^y = x^{2x} = (x^x)^2 = ((y/2)^x)^2 = (y^x)^2 \cdot 2^{-2x}$$ ...but then that's not useful unless $2^{-2x}$ is rational, which is difficult to satisfy while also satisfying $x^y \in \mathbb{Q}$.
Hint : there are positive $x,y$ such that $x^y=2$ and $y^x=3$.