If $n\mid (2^n-1)$, then $n=1$.
Somehow I am unsure if I got this right, my 'proof' seems to 'easy'. Can you please give me feedback?
So I take a prime divisor $p\mid n$. Then $p\mid (2^n-1)$, hence $2^n\equiv 1\mod p.$
So $2$ has multiplicative order $n$ in $\Bbb F_p^\times$ and therefore, by Lagrange's theorem, $n\mid (p-1)$. But since we also have $p\mid n$, this is only possible for $p=1$.