Power of 2 and divisibility by 7

Consider: $\begin{aligned} 2^0 &\equiv 1 \mod 7 \\ 2^1 &\equiv 2 \mod 7 \\ 2^2 &\equiv 4 \mod 7 \\ 2^3 &\equiv 1 \mod 7 \\ \end{aligned}$ This pattern continues. Thus $2^n\not\equiv -1\bmod 7$ for all $n$ , and $2^n + 1 \not\equiv 0\bmod 7$ . There are $\boxed{0}$ solutions.