Let's determine those prime numbers $p$ and $q$ for which $p^{q}+q^{p}$ is also a prime number.
If $p$ and $q$ are primes, then $p^{q}+q^{p}>2$, so if this sum is itself a prime, it must be odd. One of the terms, say $p^{q}$, is even, so $p=2$, and the other term is odd, so $q$ is odd.

Accordingly, we need to find those odd prime numbers $q$ for which $2^{q}+q^{2}$ is a prime number. Consider the following decomposition of this sum:

$$
2^{q}+q^{2}=\left(2^{q}+1\right)+\left(q^{2}-1\right)
$$

Since $q$ is an odd number, $2^{q}+1$ is divisible by $2+1=3$. If $q$ is not divisible by 3, then its square leaves a remainder of 1 when divided by 3, and thus $q^{2}-1$ is also divisible by 3.

We have found that if $q$ is an odd, positive integer not divisible by 3, then $2^{q}+q^{2}$ is divisible by 3. Therefore, it can only be a prime if it is exactly 3, but then $2^{q}+q^{2}=3$ would imply $q=1$, which is not a prime number.

Thus, $q$ must be divisible by 3, so the only possibility is $q=3$. Substituting, we get $2^{3}+3^{2}=17$, which is indeed a prime number.

The (unique) solution to the problem is therefore $p=2$ and $q=3$.