Why must they be positive?

Observe:

$3^m+7\equiv 0+1\equiv 1 \mod 3$ . Then, $2^n \equiv 1 mod 3$ as $2 \equiv(-1) \mod 3 \Rightarrow 2^n \equiv(-1)^n \mod 3 \therefore n=2s$ . Now we get that $3^m \equiv 4^s-7\equiv 1 \mod 4 \Rightarrow 3^m\equiv (-1)^m \mod 4 \therefore m=2t$

Now, our original equation looks like $3^{2t}+7=2^{2s}$

$7=2^{2s}-3^{2t}=(2^s-3^t)(2^s+3^t)$

$\Rightarrow 2^s-3^t=1 \text{y} 2^s+3^t=7 \Rightarrow 2^s-3^t +2^s+3^t=8 \Rightarrow 2^{s+1}=2^{3} \therefore s=2$ . Consequently, $2^{2\times2}-3^{2t}=7 \Rightarrow 3^{2t}=9 \therefore t=1$

Finally, $m=2$ y $n=4$ as unique solution $\therefore \frac{n}{m}=\frac{4}{2}=\boxed{2}$