If this claim does not hold, let's assume there exists an integer $n > 1$ such that $n\mid (2^n-1)$. Then $n$ must be an odd number because $(2^n-1)$ is odd.

Let $p$ be the least prime divisor of $n$, then $n\mid (2^n-1)$ implies $p\mid (2^n-1)$, or equivalently, $2^n\equiv 1\pmod{p}$. By Fermat's little theorem, we have $2^{p-1}\equiv 1\pmod{p}$. Let $d=gcd(n, p-1)$ then $2^d\equiv 1\pmod{p}$. By the definition of $p$, since $d\mid n$ and $d \le p-1 < p$, we get $d=1$. Then $2=2^d\equiv 1\pmod{p}$. This is a contradiction. Therefore, the previous assumption does not hold which means $(2^n-1)$ is not divisible by $n$.