Example 1 Try to determine all triples $(p, q, n)$ that simultaneously satisfy
$$
\begin{array}{l}
q^{n+2} \equiv 3^{n+2}\left(\bmod p^{n}\right), \\
p^{n+2} \equiv 3^{n+2}\left(\bmod q^{n}\right)
\end{array}
$$

where $p, q$ are odd primes, and $n$ is an integer greater than 1. ${ }^{[1]}$
(2008, China Mathematical Olympiad)
【Analysis】Obviously, $(3,3, n)(n=2,3, \cdots)$ are the arrays that meet the requirements. To prove that there are no other solutions, we can assume that there is another set of solutions that meet the requirements, and ultimately derive a contradiction.

Solution Assume $(p, q, n)$ is another set of solutions that meet the requirements. Then $p \neq q, p \neq 3, q \neq 3$.
Without loss of generality, assume $q > p \geqslant 5$.
If $n=2$, then $q^{2} \mid \left(p^{4}-3^{4}\right)$, i.e., $q^{2} \mid \left(p^{2}-3^{2}\right)\left(p^{2}+3^{2}\right)$.
By $\left(p^{2}-3^{2}, p^{2}+3^{2}\right)=\left(p^{2}-3^{2}, 2 p^{2}\right)$
$$
=\left(p^{2}-3^{2}, 2\right)=2 \text {, }
$$

we know $q^{2} \mid \left(p^{2}-3^{2}\right)$ or $q^{2} \mid \left(p^{2}+3^{2}\right)$.
But $0 < p < q$, so
$q^{n} \leqslant p^{n+2}-3^{n+2}5$.
By $p^{3} \mid \left(q^{5}-3^{5}\right)$, we get $p \mid \left(q^{5}-3^{5}\right)$.
By Fermat's Little Theorem, we know $p \nmid \left(q^{p-1}-3^{p-1}\right)$.
Therefore, $p \mid \left(q^{(5, p-1)}-3^{(5, p-1)}\right)$.
If $(5, p-1)=1$, then $p \mid (q-3)$.
Hence $\frac{q^{5}-3^{5}}{q-3}=q^{4}+3 q^{3}+9 q^{2}+27 q+81$
$\equiv 5 \times 81(\bmod p)$.
Since $p > 5$, we know $p \nmid \frac{q^{5}-3^{5}}{q-3}$. Therefore, $p^{3} \mid (q-3)$.
By $q^{3} \mid \left(p^{5}-3^{5}\right)$, we know
$$
q^{3} \leqslant p^{5}-3^{5} \quad (p \geqslant 7) \text{ and }
q^{3} \mid \left(p^{5}-3^{5}\right),
$$

we know $q^{3} \mid \frac{p^{5}-3^{5}}{p-3}$.
Thus, $q^{3} \leqslant \frac{p^{5}-3^{5}}{p-3}=p^{4}+3 p^{3}+9 p^{2}+27 p+81$.
By $5 \mid (p-1)$ and $5 \mid (q-1)$, we know $p \geqslant 11, q \geqslant 31$.
Then $q^{3} \leqslant p^{4}\left[1+\frac{3}{p}+\left(\frac{3}{p}\right)^{2}+\left(\frac{3}{p}\right)^{3}+\left(\frac{3}{p}\right)^{4}\right]$
$\left(\frac{8}{11}\right)^{\frac{1}{4}} q^{\frac{3}{4}}$
$\Rightarrow \frac{p^{5}+q^{5}-3^{5}}{p^{3} q^{3}}<\frac{p^{2}}{q^{3}}+\frac{q^{2}}{p^{3}}<\frac{1}{q}+\left(\frac{8}{11}\right)^{\frac{3}{4}} \frac{1}{q^{\frac{1}{4}}}<1$.
This contradicts $p^{3} q^{3} \mid \left(p^{5}+q^{5}-3^{5}\right)$.
In summary, $(3,3, n)(n=2,3, \cdots)$ are all the ternary arrays that meet the requirements.