We can rewrite it as $n^7+2^7=m^2+13^2$ and use the approach from USA TST 2008 P4.

I personally probably would have waited until the paper was finished being written before posting the problems online. To fill in the details for the other response, we note that $m^2 + 13^2$ is a sum of two squares, and so the only possible primes factors are $2$, prime factors of $\gcd(m, 13)$, and primes that are congruent to $1$ modulo $4$. If $2$ were one of the prime factors, then $n$ would be even. But then we would have that $n^7 - 41 \equiv 3 \pmod 4$, and so it would not be a square. Thus every prime factor of $n^7 + 2^7$ is congruent to $1$ modulo $4$. (Since $13$ is itself congruent to $1$ modulo $4$.) Since $n + 2$ is a factor of $n^7 + 2^7$, this implies that every prime factor of $n + 2$ is congruent to $1$ modulo $4$, and so $n + 2 \equiv 1 \pmod 4 \implies n \equiv -1 \pmod 4$. But then $n^7 + 2^7 \equiv (-1)^7 \equiv 3 \pmod 4$, and so $n^7 + 2^7$ would have a prime factor congruent to $3$ modulo $4$, a contradiction.

yeah same old trick as USATST 2008 p4