Example 8 Find all non-negative integer arrays $(m, n, p, q)$, such that $0<p<q,(p, q)=1$, and
$$\left(1+3^{m}\right)^{p}=\left(1+3^{n}\right)^{q} .$$
If $p>1$, by $(p, q)=1$, we can set $q=p s+r, 0<p$. Then, $\left(c^{p}-1\right) \mid\left(c^{q}-1\right)$, and
$$c^{g}-1=c^{\infty+r}-1=\left(c^{p}\right)^{s} \cdot c^{r}-1 \equiv 1^{s} \cdot c^{r}-1\left(\bmod c^{\rho}-1\right),$$

Therefore,
$$\left(c^{p}-1\right) \mid\left(c^{r}-1\right)$$

This cannot hold when $r<p$, so $p=1$, and (10) becomes
$$1+3^{m}=\left(1+3^{n}\right)^{q}$$

Noting that $q>p=1$, hence $q \geqslant 2$, at this time the right side of (11) $\equiv 0(\bmod 4)$, so
$$3^{m} \equiv-1(\bmod 4)$$

Thus, $m$ is odd, at this moment
$$1+3^{m}=(1+3)\left(1-3+3^{2}-\cdots+3^{m-1}\right)=4 \cdot A,$$

Here $A=1-3+3^{2}-\cdots+3^{m-1}$ is the sum of $m$ odd numbers, so $A$ is odd, thus $(4, A)=1$. This requires 4 and $A$ to both be $q$-th power numbers, from which we can deduce $q=2$, and then we can find
$$(m, n, p, q)=(1,0,1,2),$$

It is easy to verify that it meets the requirements.