Are there only two solutions for $$\sum_{k=0}^m3^k=2^n$$

Such as $3^0=2^0$ and $3^0+3^1=2^2$

Note

• If $m$ is even then $\sum_{k=0}^m3^k$ will be odd.

• $$\sum_{k=0}^m3^k=\sum_{k=0}^m\binom{m+1}{k+1}2^k=\sum_{k=0}^{m}\sum_{l=0}^{k}\binom{m+1}{k+1}\binom{k}{l}$$

Are there only two solutions for $$\sum_{k=0}^m3^k=2^n$$

Such as $3^0=2^0$ and $3^0+3^1=2^2$

Note

• If $m$ is even then $\sum_{k=0}^m3^k$ will be odd.

• $$\sum_{k=0}^m3^k=\sum_{k=0}^m\binom{m+1}{k+1}2^k=\sum_{k=0}^{m}\sum_{l=0}^{k}\binom{m+1}{k+1}\binom{k}{l}$$

Edit: generalization may be more interesting (motivation).

We can also ask as, Are there finitely many pairs of $(m, n)$ for any positive integer $(x, y)$ greater than $1$ such that $1+x+x^2+...+x^m=y^n$ Or $x^{m+1}-2\cdot y^{n}=1$