Find n

Since $2^3 \equiv 8 \equiv 1 \pmod 7 \implies 2^n \bmod 7 = \begin{cases} 1 & \text{if }n \mod 3 = 0 \\ 2 & \text{if }n \mod 3 = 1 \\ 4 & \text{if }n \mod 3 = 2 \end{cases}$

$\implies \left(2^n + 1\right) \bmod 7 = \begin{cases} 2 & \text{if }n \mod 3 = 0 \\ 3 & \text{if }n \mod 3 = 1 \\ 5 & \text{if }n \mod 3 = 2 \end{cases}$

Therefore, there is $\boxed{0}$ $n$ such that $7 \mid 2^n + 1$ .