Well-known (as well as the method to solve it). Let $p$ be the least prime dividing $n$, so a fortiori $p\mid 3^n - 2^n$. Thus $p>3$. We then have $3^n \equiv 2^n \pmod{p}$, or $(3\cdot2^{-1})^n \equiv 1 \pmod{p}$. Let $\nu$ be the multiplicative order of $3\cdot2^{-1}$ modulo $p$; we then must have $\nu \mid n$ and also $\nu \mid p-1$ (by Fermat's little). But then $\nu \leq p-1<p$, and so from $\nu \mid n$ follows $\nu = 1$ by the minimality of $p$. That means $3\cdot2^{-1} \equiv 1 \pmod{p}$, i.e. $3 \equiv 2 \pmod{p}$, absurd.

Suppose there exist an integer $n$ such that $n>1$ and $n|3^n-2^n.$ Clearly $n$ is an odd integer. Let $p$ be the smallest prime divisor of $n$.Hence $p\geq5.$ Since $p$ is odd,so there exist an integer $k$ such that $2k\equiv1 (mod p)$ $p|3^n-2^n \implies p|(3k)^n-(2k)^n \implies (3k)^n\equiv1 (mod p)$ $gcd(3k,p)=1 \implies (3k)^{p-1}\equiv1(mod p)$ So $p|(3k)^{{gcd(p-1,n)}}-1 \implies p|3k-1$ So $p|3(2k-1)-2(3k-1) \implies p|1$ which is impossible. So there is no integer $n$ such that $n>1$ and $n|3^n-2^n.$