Can an odd number $n$ divide $2^n-1$?

Clearly an even number $n$ cannot divide $2^{n}-1$, but about odd ones ? If $n$ is an odd prime this cannot happen neither since for an odd prime $p$ we have $2^p\equiv 2\pmod p$ and so $p$ cannot divide $2^p-1$ but what about the general case ($i.e~n$ an odd number larger than 1) ?

The only (odd) value for $n$ for which $n \mid 2^n - 1$ is $n=1$.

Suppose for the sake of contradiction that $n>1$ is such that $n \mid 2^n - 1$ and let $p$ be the smallest prime divisor of $n$. Then we have $p \mid n \mid 2^n - 1$ or $2^{n} \equiv 1 \mod p$. We also have $2^{p-1} \equiv 1 \mod p$ by Fermat's little theorem. It follows that $2^{\gcd(n,p-1)} \equiv 1 \mod p$. Note that $n$ and $p-1$ are coprime since $p$ is the smallest prime divisor of $n$. Thus $2^1 \equiv 1 \mod p$, contradiction.

Hint $ $ mod $\rm\color{#c00}{least}$ prime $\,p\mid n\!:\ 2^n \equiv 1\Rightarrow\, 2\,$ has order $\,k\mid n\,\color{#c00}{\Rightarrow}\ k \ge$ $\,p\,\Rightarrow\, 2^{p-1}\not\equiv 1\,\Rightarrow\!\Leftarrow$

Note $ $ Key Idea is: $ $ if $\ a\not\equiv 1,\,\ a^n\equiv 1\,$ then the order of $\,a\,$ is $\ge$ least prime $\,p\mid n.$