What are the natural solutions $x, c$ to $2^x + 1 = c\cdot 3^2$ for odd $x$?

Question

I understand that $x=3, c=1$ is a solution. Are there any more natural number solutions? If so, how would I find them? If not, why not?

Answer 1

Using mod $9$, and you will have: $2^1 = 2, 2^2 = 4, 2^3 = 8, 2^4 = 7, 2^5 = 5, 2^6 = 1, 2^7 = 2$. Thus the only solution is $x = 3k$ where $k$ is odd. So the solution set is: $S = \{3,9,15,21,...\}$.