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. ${ }^{[1]}$
Assume $(p, q, n)$ is another set of solutions. Then $p \neq q, p \neq 3, q \neq 3$.
Without loss of generality, let $q > p \geqslant 5$.
If $n=2$, then $q^{2} \mid \left(p^{4}-3^{4}\right)$, which means $q^{2} \mid \left(p^{2}-3^{2}\right)\left(p^{2}+3^{2}\right)$.
Since $\left(p^{2}-3^{2}, p^{2}+3^{2}\right)=\left(p^{2}-3^{2}, 2 p^{2}\right)$
$$
=\left(p^{2}-3^{2}, 2\right)=2 \text {, }
$$

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

we know $q^{3} \mid \frac{p^{5}-3^{5}}{p-3}$.
Thus, $q^{3} \leqslant \frac{p^{5}-3^{5}}{p-3}=p^{4}+3 p^{3}+9 p^{2}+27 p+81$.
Since 5 divides $(p-1)$ and 5 divides $(q-1)$, 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 conclusion, $(3,3, n) (n=2,3, \cdots)$ are all the solutions that meet the requirements.