There are two ways to try to prove this. One is in the title, the other is its de Morgan counterpart: $n \nmid 2^{2^{2^n+1}+1}+1 \implies n \nmid 2^{2^n+1}+1$. Disproving it requires only one example of course.

Tried using $\gcd(2^a+1, 2^b+1) = 2^{\gcd(a, b)}+1$ (where $a$ and $b$ are odd positive integers), stuck on both ends. I figured out that if $n$ divides $2^n+1$ then n divides both $2^{2^n+1}+1$ and $2^{2^{2^n+1}+1}+1$ but this implication doesn't work backwards (e.g. $n=57$).

Would appreciate some help.

EDIT1 Eric's pointer wasn't enough for me. Trying to bump by editing instead of reposting (sorry, not sure how to).

EDIT2 This is not much but might save some time for someone. Using user101140's notation and the $a^n+b^n$ identity

$f(n) = 2^n+1 = 3 \sum\limits_{k=0}^{n-1} (-2)^k$

$f(f(n)) = 3 \sum\limits_{k=0}^{2^n} (-2)^k$

$f(f(f(n))) = 3 \sum\limits_{k=0}^{2^{2^n+1}} (-2)^k$

Also, $n \mid f(n) \implies n \mid f(f(n))$ is due to $n \mid f(n) \implies f(n) \mid f(f(n))$ so the proof might be something along the lines of $n \mid f(f(n)) \implies f(f(n)) \mid f(f(f(n)))$. (Please don't bash me if that's stupid.)