For some positive integers $m$ and $n, 2^{m}-2^{n}=1792$. Determine the value of $m^{2}+n^{2}$.
Since $2^{11}=2048$ and $2^{8}=256$, then $2^{11}-2^{8}=2048-256=1792$.

Therefore, $m=11$ and $n=8$, which gives $m^{2}+n^{2}=11^{2}+8^{2}=121+64=185$.

Alternatively, we can factor the left side of the equation $2^{m}-2^{n}=1792$ to obtain $2^{n}\left(2^{m-n}-1\right)=$ 1792 .

Since $m>n$, then $2^{m-n}-1$ is an integer.

Now, $1792=2^{8} \cdot 7$.

Since $2^{n}\left(2^{m-n}-1\right)=2^{8} \cdot 7$, then $2^{n}=2^{8}$ (which gives $n=8$ ) and $2^{m-n}-1=7$ (which gives $m-n=3$ and so $m=11$ ).

ANSWER: 185