Here is a baby answer.

**Theorem:** There exist two irrational numbers $p,q$ such that $p^q$ is rational.

*Proof:* In case $\sqrt 2^{\sqrt 2}$ is rational we can take $p=q=\sqrt 2$.

Otherwise take $p= \sqrt 2^{\sqrt 2}$ and $q=\sqrt 2$. We have $p^q = \big(\sqrt 2^{\sqrt 2}\big)^{\sqrt 2} = \sqrt 2^{\sqrt 2 \cdot \sqrt 2} = \sqrt 2^2=2$ is rational.