Clearly, $p \ne q$. So we may assume that $p < q$ without loss of generality. Assume that $p = 2$. Then since
$$q^q + 5 \equiv 5 \equiv 0 \pmod{q},$$
the only possible prime for $q$ is $5$. Furthermore, $(p, q) = (2, 5)$ satisfies the above condition. Now we assume that both $p$ and $q$ are odd primes. Since $p^p + 1 \equiv 0 \pmod{q}$, $p^{p-1} - p^{p-2} + \dots - p + 1$ is divisible by $q$ and $p^{2p} \equiv 1 \pmod{q}$. On the other hand, Fermat's Little Theorem says
$$p^{q-1} \equiv 1 \pmod{q}.$$
If $\gcd(2p, q-1) = 2$, then $p^2 \equiv 1 \pmod{q}$. Hence $p \equiv 1 \pmod{q}$ or $p \equiv -1 \pmod{q}$. Therefore
$$0 \equiv p^{p-1} - p^{p-2} + \dots - p + 1 \equiv 1 \pmod{q},$$
which is a contradiction. Assume that $\gcd(2p, q - 1) = 2p$, that is, $q \equiv 1 \pmod p$. In this case we have
$$0 \equiv p^p + q^q + 1 \equiv p^p + 1 + 1 \equiv 2 \pmod{p},$$
which is again a contradiction.
Therefore $(p, q) = (2, 5)$ or $(5, 2)$ is the only pair of primes satisfying the above condition. $\square$