Example 11 Find all positive integers $m, n$, such that $2^{m}+3^{n}$ is a perfect square.

Find all positive integers $m, n$, such that $2^{m}+3^{n}$ is a perfect square.
Solution: Let $2^{m}+3^{n}=k^{2} \quad\left(k \in \mathbb{Z}^{+}\right)$
(5)
It is easy to see that $2 \nmid k, 3 \nmid k$. Taking the modulus 3 of both sides of (5) yields $(-1)^{m} \equiv 1(\bmod 3)$, hence $m$ is even.

Let $m=2 m_{1}\left(m_{1} \in \mathbb{Z}^{+}\right)$, then $2^{m}=4^{m_{1}}$. Taking the modulus 4 of both sides of (5) yields $(-1)^{n} \equiv k^{2}(\bmod 4)$.
Since the remainder of a square number modulo 4 cannot be -1, $n$ must also be even.
Let $n=2 n_{1}\left(n_{1} \in \mathbb{Z}^{+}\right)$, then (5) becomes $\left(k-3^{n_{1}}\right)\left(k+3^{n_{1}}\right)=2^{2 m_{1}}$. Therefore, $k-3^{n_{1}}$ and $k+3^{n_{1}}$ are both non-negative integer powers of 2, but $\left(k-3^{n_{1}}\right)+\left(k+3^{n_{1}}\right)=$ $2^{k}$, which is not a multiple of 4, so $k-3^{n_{1}}=2, k+3^{n_{1}}=2^{2 m_{1}-1}$, thus $2^{2 m_{1}-2}=3^{n_{1}} +1$, i.e., $\left(2^{m_{1}-1}-1\right)\left(2^{m_{1}-1}+1\right)=3^{n_{1}}$.

Since two consecutive odd numbers are coprime, we have $2^{m_{1}-1}-1=1$, i.e., $m_{1}=2$, and $n_{1}=1$, hence $m=4, n=2$.