If $p=q\neq 2$, then $p^{q^2-q+1}+q^{p^2-p+1}-p-q = 2p(p^{p^2-p}-1)$ which is a multiple of $p$ but not $p^2$, so that it isn't a perfect square. $p=q=2$ is also impossible since $p^{q^2-q+1}+q^{p^2-p+1}-p-q = 12$ isn't a perfect square. Otherwise, then note that $\phi(p^2) = p^2-p$ so that $p^2\mid q^{p^2-p}-1$ which implies \[ p^{q^2-q+1}+q^{p^2-p+1}-p-q = p(p^{q^2-q}-1) + q(q^{p^2-p}-1) \]is a multiple of $p$ but not $p^2$. Conclusion follows.

A solution with $FLT$ Assume ${p}^{{q}^{2} - q + 1} + {q}^{{p}^{2} - p + 1} - p - q = {a}^{2}$ and $p > q$ so $gcd(p,q) = 1$ so $ {p}^{q-1} \equiv 1 \pmod{q}$ so $ {p}^{{q}^{2} - q} \equiv 1 \pmod{q}$ so $ {p}^{{q}^{2} - q + 1} \equiv p \pmod{q}$ and therefore ${p}^{{q}^{2} - q + 1} + {q}^{{p}^{2} - p + 1} - p - q \equiv p - p \equiv 0 \pmod{q}$ so we get that $q$ divides ${a}^{2}$ so $ a \equiv 0 \pmod{q}$ hence $ p^{q^2 - q +1} - p - q \equiv 0 \pmod{q^2}$ but we know that $ p^{q^2 - q +1} - p \equiv 0 \pmod{q^2}$ so $q^2$ divides $q$ which is contradiction. The case $p$ = $q$ can be checked easily.