Example 9 Find all prime numbers $p, q, r$ that satisfy $p^{q}+q^{p}=r$.


Solution: Clearly, $p \neq q$. Without loss of generality, assume $p<q$.
$\because r$ is a prime number,
$\therefore p$ and $q$ cannot both be odd.
Thus, $p=2$.
When $q$ is a prime number not less than 5, we have
$$
\begin{array}{l}
p^{q}+q^{p}=2^{q}+q^{2} \\
=\left(2^{q}+1\right)+\left(q^{2}-1\right) \\
=3 \times \frac{2^{q}+1}{2+1}+\left(q^{2}-1\right) \\
=3\left(2^{q-1}-2^{q-2}+\cdots+1\right)+(q+1)(q-1) .
\end{array}
$$

Since $q$ is not 3 and is not a multiple of 3, among the three consecutive natural numbers $q-1$, $q$, and $q+1$, there must be one that is a multiple of 3, hence $(q-1)(q+1)$ is a multiple of 3.

Therefore, $p^{q}+q^{p}$ is a multiple of 3, and cannot be a prime number, which is a contradiction. Thus, $q<5$.
Therefore, the only primes satisfying $p^{q}+q^{p}=r$ are
$$
p=2, q=3, r=17 \text {. }
$$