Determine all pairs $(p, q)$ of positive integers such that $p$ and $q$ are prime, and $p^{q-1}+q^{p-1}$ is the square of an integer.
If $(p, q)=(2,2)$, then $p^{q-1}+q^{p-1}=2^{1}+2^{1}=4$, which is indeed the square of an integer. We will prove that this is the only pair of positive integers that satisfies the conditions of the problem.

If $p$ and $q$ are odd, then $p^{q-1}$ and $q^{p-1}$ are squares of odd integers, so we have
$$
p^{q-1} \equiv q^{p-1} \equiv 1 \quad(\bmod 4) \quad \Rightarrow \quad p^{q-1}+q^{p-1} \equiv 2 \quad(\bmod 4) .
$$

Since the square of an integer must be congruent to 0 or 1 modulo 4 , there are no pairs $(p, q)$ that satisfy the conditions of the problem when $p$ and $q$ are odd.

We have deduced that a pair $(p, q)$ that satisfies the conditions of the problem must have $p=2$ or $q=2$. Without loss of generality, let us assume that $q=2$, so that we are looking for a prime $p$ and a positive integer $m$ such that $p+2^{p-1}=m^{2}$. We have already dealt with the case $p=2$, so let us henceforth assume that $p$ is odd. Then we may write this equation as
$$
\left(m+2^{\frac{1}{2}(p-1)}\right)\left(m-2^{\frac{1}{2}(p-1)}\right)=p .
$$

Hence, we require $p=m+2^{\frac{1}{2}(p-1)}>2^{\frac{1}{2}(p-1)}$. However, we will prove below that the inequality $p>2^{\frac{1}{2}(p-1)}$ fails to be true for $p \geq 7$. So it remains to check $p=3$ and $p=5$, but these two cases do not yield solutions. Therefore, we conclude that $(p, q)=(2,2)$ is the only pair that satisfies the conditions of the problem.
For completeness, we now prove that $n \leq 2^{\frac{1}{2}(n-1)}$ for all $n \geq 7$ via induction. For $n=7$, one can check the claim directly. Now if the claim is true for some $n \geq 7$, then it is also true for $n+1$ since
$$
n<2^{\frac{1}{2}(n-1)} \quad \Rightarrow \quad n+1<\frac{n+1}{n} \times 2^{\frac{1}{2}(n-1)} \leq \frac{8}{7} \times 2^{\frac{1}{2}(n-1)}<\sqrt{2} \times 2^{\frac{1}{2}(n-1)}=2^{\frac{1}{2} n} .
$$

This completes the induction argument.

Comments
One can consider the same problem, but relax the condition that $p$ and $q$ are prime. A computer search with $1 \leq p, q \leq 100$ only reveals solutions with $p=q$. Thus, we are left with the following problem: if $p$ and $q$ are positive integers such that $p^{q-1}+q^{p-1}$ is the square of an integer, is it necessarily true that $p=q$ ?