Example 4 Find all prime triples $(p, q, r)$ such that
$$
p \text { | }\left(q^{r}+1\right), q\left|\left(r^{p}+1\right), r\right|\left(p^{q}+1\right) \text {. }
$$
Given that $p, q, r$ are pairwise coprime, we know they are distinct.
(1) $p, q, r$ are all odd primes.

From $p \mid (q^r + 1)$, we get $\delta_p(q) = 2$ or $2r$.
If $\delta_p(q) = 2r$, then
$$
\begin{array}{l}
2r \mid (p-1) \Rightarrow p \equiv 1 \pmod{r} \\
\Rightarrow p^q + 1 \equiv 2 \pmod{r},
\end{array}
$$
which is a contradiction.
Thus, $\delta_p(q) = 2 \Rightarrow p \mid (q+1)$.
Since $p, q$ are both odd, we have $p \mid \frac{q+1}{2}$.
Therefore, $p \leq \frac{q+1}{2} < q$.
Similarly, $q < r, r < p$. This is a contradiction.
(2) Among $p, q, r$, there is a 2.

By cyclic symmetry, assume $p = 2$.
From $r \mid (p^q + 1)$, similarly, we get $r \mid (p+1)$.
Thus, $r = 3$.
Then, from $q \mid (3^2 + 1)$, we get $q = 5$.
Therefore, all solutions are
$$
(p, q, r) = (2, 5, 3), (5, 3, 2), (3, 2, 5)
$$