<s>[INST] Prove that $2^x = 3 \cdot 9^m+5$ has no positive integer solutions for $m \geq 2$ 
Prove that $2^x = 3 \cdot 9^m+5$ has no positive integer solutions for $m \geq 2$. 

I noticed that $x \equiv 5 \bmod 6$ and thus $2^x \equiv 4 \bmod 7$, but that doesn't seem to help me since $3 \cdot 9^4 +5 \equiv 4 \bmod 7$. Pretty much any other mod I use doesn't seem to work so I think proof by contradiction or something may work better.
 [/INST] Suppose that $x$ is a solution, then:
$ 2^{x} = 3.9^{m} + 5 = 3.9^{m} + 3 + 2$, so:
$$2(2^{x-1} - 1) = 3( 9^{m} + 1) \Rightarrow (3.(9^{m} + 1)) | (2^{x-1} - 1)$$
For $m\geq 2$. But $2^{x-1} - 1$ is odd and $3.(9^{m} + 1)$ is pair.
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