Suppose that $\sqrt{7+\sqrt{48}}=m+\sqrt{n}$. Squaring both sides, we obtain $7+\sqrt{48}=(m+\sqrt{n})^{2}$. Since $(m+\sqrt{n})^{2}=m^{2}+2m\sqrt{n}+n$, then $7+\sqrt{48}=(m^{2}+n)+2m\sqrt{n}$. Let's make the assumption that $m^{2}+n=7$ and $2m\sqrt{n}=\sqrt{48}$. Squaring both sides of the second equation, we obtain $4m^{2}n=48$ or $m^{2}n=12$. So we have $m^{2}+n=7$ and $m^{2}n=12$. By inspection, we might see that $m=2$ and $n=3$ is a solution. If we didn't see this by inspection, we could note that $n=7-m^{2}$ and so $m^{2}(7-m^{2})=12$ or $m^{4}-7m^{2}+12=0$. Factoring, we get $(m^{2}-3)(m^{2}-4)=0$. Since $m$ is an integer, then $m^{2} \neq 3$. Thus, $m^{2}=4$ which gives $m=\pm 2$. Since $m$ is a positive integer, then $m=2$. When $m=2$, we get $n=7-m^{2}=3$. Therefore, $m=2$ and $n=3$, which gives $m^{2}+n^{2}=13$. We note that $m+\sqrt{n}=2+\sqrt{3}$ and that $(2+\sqrt{3})^{2}=4+4\sqrt{3}+3=7+4\sqrt{3}=7+\sqrt{48}$, as required. This means that, while the assumption we made at the beginning was not fully general, it did give us an answer to the problem.