7. Find all prime triples $(p, q, r)$ such that
$$p \mid \left(q^{r}+1\right), q \mid \left(r^{p}+1\right), r \mid \left(p^{q}+1\right) .$$
7. Let $(p, q, r)$ be an array of prime numbers, and by cyclic symmetry, assume $p = \max \{p, q, r\}$. From $p \mid (q^r + 1)$, we know $p \neq q$, similarly $q \neq r$, $r \neq p$, i.e., $p, q, r$ are pairwise distinct. We discuss the following two cases:
(1) $p > q > r$. From $q \mid (r^p + 1)$, i.e., $r^p \equiv -1 \pmod{q}$, we know $r^{2p} \equiv 1 \pmod{q}$, hence $\delta_q(r) \times p$ but $\delta_q(r) \mid 2p$. By Fermat's Little Theorem, we know $r^{q-1} \equiv 1 \pmod{q}$, hence $\delta_q(r) \mid (q-1)$. Since $q-1 < r$, we have $\delta_q(r) \mid 2$. 

By $p \mid (q^r + 1)$, similarly, we get $\delta_p(q) \mid 2r$. Since $p \geq r + 2 > q + 2$, we have $\delta_p(q) > 2$, thus $\delta_p(q) = 2r$. Combining Fermat's Little Theorem, we get $2r \mid (p-1)$, hence $p \equiv 1 \pmod{r}$, leading to
$$0 \equiv p^q + 1 \equiv 2 \pmod{r},$$

i.e., $r \mid 2$, a contradiction. $\square$
In summary, $(p, q, r) = (5, 3, 2), (3, 2, 5)$, or $(2, 5, 3)$.