What exactly are those "two irrational numbers" $x$ and $y$ such that $x^y$ is rational?

Let $x=3^{1/2}$ and $y=\log_{3}(4)$. Then $x^y=2$.

The proof that $x$ is irrational is familiar. For $y$, suppose $y=p/q$ where $p$ and $q$ are positive integers. Then $3^{p/q}=4$, so $3^p=4^q$. This is impossible, since $4^q$ is even and $3^p$ is odd.

Let $x = \mathrm{e}$ and $y = \ln(2)$, then $x^y = \mathrm{e}^{\ln(2)} = 2$.