$p (p^2 + 1) = q (q + 1)$. It is clear that $q > p$, because otherwise we have inequality. Thus $p \nmid q$, so $p | q + 1$. Also $q | p^2 + 1$. Thus $pq | p^2 + q + 1$. But since $q > p$, we have $2pq > pq + p^2 > q + 1 + p^2$. Thus $pq = p^2 + q + 1$, so $q(p-1) = p^2 + 1$. This means $p - 1 | p^2 + 1$, so $p - 1 | 2$, or $p = 2, 3$. Answer $(3, 5)$

How can $p | q + 1$?

As $p$ and $q$ is always odd except $p=2$ or $q=2$

An odd can divide an even

oh yes you are right.I was a bit confused.

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Can we solve this problem by Vieta Jumping?