There are two irrational numbers \(x\) and \(y\) such that \(x^y\) is rational.
Proof:
broke:
Either \(\sqrt{2}^\sqrt{2}\) is rational, or if not, then \((\sqrt{2}^\sqrt{2})^\sqrt{2} =2\) is rational, but we don't know which case holds, hence one of \(x=y=\sqrt{2}\) or \(x=\sqrt{2}^\sqrt{2}\) and \(y=\sqrt{2}\) gives what we want.
woke:
By the really hard Gelfond–Schneider theorem we know \(x=\sqrt{2}^\sqrt{2}\) is irrational, and \(y=\sqrt{2}\) is irrational by an elementary proof, and \(x^y=2\), so these are the numbers we want.
bespoke:
By elementary proofs we know that both \(x=\sqrt{2}\) and \(y=\log_23\) are irrational, and \(x^{2y}=3\).