Let $n$ be a positive integer such that
$$p^{q-1} + q^{p-1} = n^2.$$
We will divide our solution into cases, depending on the parity of numbers $p$ and $q$.

If both are even, the only possibility is that both are equal to $2$. This leads to a solution $(p, q) = (2, 2)$ for $n = 2$.

If both $p$ and $q$ are odd, we conclude that there are no solutions by looking at the remainder of division by $4$. If $p$ and $q$ are odd, then $p-1$ and $q-1$ are even, so that $p^{q-1}$ and $q^{p-1}$ are squares of odd numbers. This means that they always give remainder $1$ when divided by $4$. This shows that the left-hand side gives remainder $2$ when divided by $4$, whereas $n^2$ must give remainder $1$ or $0$ when divided by $4$.

The only remaining case is when one of the numbers $p$ and $q$ is even, while the other is odd. The equation being symmetric, we can, without loss of generality, assume that $p$ is odd, i.e. $p = 2k+1$ for some positive integer $k$, and that $q$ equals $2$.

Now we have
$$p + 2^{p-1} = n^2, \quad \text{ i.e. } p = n^2 - 2^{2k} = (n - 2^k)(n + 2^k).$$
Since $p$ is prime, the only way in which we can represent it as a product $(n - 2^k)(n + 2^k)$ is if one of the factors equals $\pm 1$, while the other is $\pm p$. Notice that $n + 2^k > 0$, and also that $n + 2^k > n - 2^k$, which yields
$$n - 2^k = 1 \quad \text{and} \quad n + 2^k = p.$$
Subtracting these two equalities, we get
$$2^{k+1} + 1 = p = 2k + 1.$$
Let us prove that this equation has no solution. We know that $2^{k+1} + 1$ gives remainder $1$ when divided by $4$, so the same must hold for $2k + 1$. This implies that $k$ is even, i.e. $k = 2l$ for some positive integer $l$. In that case,
$$p = 2^{k+1} + 1 = 2^{2l+1} + 1 = 2 \cdot 4^l + 1$$
is divisible by $3$ (because $4^l$ gives remainder $1$ when divided by $3$). We conclude that $p$ is divisible by $3$, so $p = 3$. Plugging this into the initial equation, we see that $p = 3$ and $q = 2$ is not a solution.

This means that the only solution is $(p, q) = (2, 2)$.