Solution:
Without loss of generality, assume that $a$ is even and consequently $b$ is odd. Let $a = 2 a'$. Then
$$2 a^{p} b - 2 a b^{q} = 4 a' b \left(a^{p-1} - b^{q-1}\right)$$
If this is a square, then $a'$, $b$ and $a^{p-1} - b^{q-1}$ are pairwise coprime.

On the other hand, $a^{p-1}$ is divisible by $4$ and $b^{q-1}$ gives the remainder $1$ when divided by $4$. It follows that $a^{p-1} - b^{q-1}$ has the form $4k + 3$, a contradiction.