Let $n=j\cdot 2^k$ where $j$ is an odd number and $k$ is a non-negative integer. If $j>1$, we have $$2^n + 1 = \big(2^{\ j\cdot 2^k}\big) + 1^j = \big(2^{\ 2^k}\big)^j + 1^j =\big(2^{2^k}+1\big)(\cdots)$$ where $(\cdots)$ is a polynomial of $2^{2^k}$. It is clear that both $\big(2^{2^k}+1\big)$ and $(\cdots)$ are greater than $1$. This implies $2^n+1$ is a composite number. Therefore, if $(2^n+1)$ is a prime number, $j$ must be 1 which means $n$ is some power of $2$.