Prove that there exist two irrational numbers x and y such that xy is rational

From a more abstract point of view, suppose you have a set $X$, a subset $\mathcal{I} \subset X$, an element $a \subset \mathcal{I}$ and a map $b : X \to X$ satisfying $b^n(a) \not\in \mathcal{I}$ for some $n$.

Then necessarily there exists $m$ such that $b^m(a) \in \mathcal{I}$ and $b^{m+1}(a) \not\in \mathcal{I}$.

The proof you refer to is not really mysterious; it's only presented to look mysterious, specifically by the choice of $\sqrt{2}$. There are plenty of choices for irrational $a$ and $b$, such that $a^{b^n}$ is rational for some $n$, and exactly the same conclusion can be drawn.

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It's possible to prove nonconstructively that there exists irrational numbers $x$ and $y$ such that $x^y$ is rational, but that proof only proves that such numbers exist and does not specify what they are.

What is a constructive proof that there are two irrational numbers $x$, $y$ such that $x^y$ is rational, i.e. what are those numbers?