Clearly, $n$ is odd because $(2^n+1)$ is odd. Let $p$ be the minimal prime divisor of $n$. We will show that $p=3$ which implies $3\mid n$.

Let $r$ be the order of $2$ modulo $p$, then $$2^r\equiv 1\pmod{p}$$

Because $n\mid(2^n+1)$ and $p\mid n$, we have $$2^n\equiv -1\pmod{p} \implies 2^{2n}=(2^n)^2\equiv 1\pmod{p}$$

Meanwhile, because $p\ge 3$ is odd, $2^{p-1}$ cannot be divisible by $p$. Applying Fermat's little theorem leads to $$2^{p-1}\equiv 1\pmod{p}$$

Because $r$ is the order of 2 modulo $p$, the previous three relations imply $r\mid 2n$ and $r\mid(p-1)$. Therefore we conclude $r\mid (2n, p-1)$.

Now we show that $(2n, p-1)=2$. If so, $r\mid 2\implies r=2$. In this case the first equation above will lead to the desired result of $p=3$.

Because $p$ is odd, therefore $(p-1)$ is even. This means that $2\mid (2n, p-1)$. However, because $n$ is odd, $4 \not\mid 2n$ which implies $4\not\mid (2n, p-1)$.

Now, we claim that no odd prime can divide $(2n, p-1)$. This is because assuming there exists such an odd prime $q$, then $q\mid 2n$ implies $q\mid n$. Meanwhile $q\mid (p-1)$ implies $q < p$. This contradicts to the earlier assumption $p$ is the smallest prime divisor of $n$.

Therefore, $(2n, p-1)$ is a multiple of $2$, but not $4$ or any odd prime. This means $(2n, p-1)=2$.