Without loss of

Let $a=2 a^{\prime}$. If $\vdots$. Without loss of generality, assume that $a$ is even and consequently $b$ is odd.

$$2 a^{p} b-2 a b^{q}=4 a^{\prime} b\left(a^{p-1}-b^{q-1}\right)$$

is a square, then $a^{\prime}, b$ 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 $4 k+3$, a contradiction.