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



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4. Find all pairs of positive integers $m, n$ such that $2^{m}+3^{n}$ is a perfect square.
4. Let $2^{m}+3^{n}=x^{2}$, then 3 divides $x$ and $x$ is odd, so $x^{2} \equiv 1(\bmod 3)$, so $2 \mid m$, so $m \geqslant 2$ which means $4 \mid 2^{m}$.

Also, $x^{2} \equiv 1(\bmod 4) \Rightarrow 2 \mid n$, so, let $m=2 p, n=2 q$ we have $2^{2 p}+3^{2 q}=x^{2}$, thus
$$
2^{2 p}=\left(x-3^{q}\right)\left(x+3^{q}\right) \text {. }
$$

Also, $\left(x-3^{q}, x+3^{q}\right)=2$, so $x+3^{q}=2^{2 p-1}, x-3^{q}=2$.
Also, $3^{2 q}=\left(x-2^{p}\right)\left(x+2^{p}\right)$
and $\left(x-2^{p}, x+2^{p}\right)=1$.

So $x-2^{p}=1, x+2^{p}=3^{2 q}$.
In summary, $x=5$, so there is a unique pair of positive integers $m=4, n=2$.