7. Find all positive integers $m, n$, such that $2^{m}+3^{n}$ is a perfect square.
7. Solution: Let $2^{m}+3^{n}=k^{2}\left(k \in \mathbf{N}_{+}\right)$.
(1)
(1) Taking both sides modulo 3, since $3 \mid k$, then $k^{2} \equiv 1(\bmod 3)$, hence $(-1)^{m} \equiv 1(\bmod 3)$, which implies $2 \mid m, m \geqslant 2$.
(1) Taking both sides modulo 4, since $k$ is odd, $k^{2} \equiv 1(\bmod 4)$, hence $(-1)^{n} \equiv 1(\bmod 4)$, which implies $2 \mid n$.

Let $m=2 m_{1}, n=2 n_{1}\left(m_{1}, n_{1} \in \mathbf{N}_{+}\right)$, then $3^{2 n_{1}}=k^{2}-2^{2 m_{1}}=\left(k+2^{m_{1}}\right)\left(k-2^{m_{1}}\right)$.
Since $\left(k+2^{m_{1}}\right)-\left(k-2^{m_{1}}\right)=2^{m_{1}+1} \neq 0(\bmod 3)$, one of $k+2^{m_{1}}, k-2^{m_{1}}$ must not be a multiple of 3, but they are both multiples of 3, so it must be that $k-2^{m_{1}}=1, k+2^{m_{1}}=2^{m_{1}+1}+1=3^{2 n_{1}}$.

Thus, $2^{m_{1}+1}=3^{2 n_{1}}-1=\left(3^{n_{1}}+1\right)\left(3^{n_{1}}-1\right)$, and $\left(3^{n_{1}}+1,3^{n_{1}}-1\right)=\left(3^{n_{1}}+1,2\right)=2$, so it must be that $3^{n_{1}}+1=2^{m_{1}}, 3^{n_{1}}-1=2$, hence $n_{1}=1, m_{1}=2 \Rightarrow n=2, m=4$.
In conclusion, the solution $(m, n)$ is $(4,2)$.