How to prove there exists at least one positive integer solution of $k$ and $n$ in $2^{k} = 3^{n-1}(m + 1) - 1$, where m is any even integer?

Question

Here's the related question

Is it possible to prove there exists at least one positive integer solution for $k$ and $n$ in $2^{k} = 3^{n-1}(m + 1) - 1$, where m is any even positive integer?

So we need to prove, that for any given even positive integer m, there exists at least one positive integer solution of k and n.

Eg:
If m = 2, then k = 3, n = 2
If m = 4, then k = 2, n = 1

Answer 1

This is still false as stated. With $m=26$ we have the equation $$2^k=3^{n+2}-1.$$ But by Catalan's conjecture (now a theorem of Mihăilescu) the only non-trivial consecutive integer powers are $2^3$ and $3^2$ in violation of the requirement that $n$ should be a positive integer.

The same argument works whenever $m+1$ is a power of $3$ larger than $9$.

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This observation also suggests that for a given fixed $m$ the problem may be rather difficult.

Answer 2

The problem is wrong, take $m=-10$, we have: $2^{k} = 3^{n}(-19) - 1$ As $3^{n}\geq0 $, we have $-19.3^n\leq0$ and trivially $-19.3^n-1<0$ which is equal to $2^k$, so $2^k<0$, contradiction.

You should strict $m$ to be a positive integer.

Answer 3

I rewrite this: $$ {2^k+1 \over m+1} = 3^n \tag 1 $$

1. ) If $k$ is odd, then the numerator in the lhs contains $3$ to some power $n$, so let $m+1$ be the cofactor (aside of powers of $3$) to have a solution.
2. ) if $k$ is even, then no solution is possible.

Observation 1) gives an infinite set of solutions for $m+1$, but this set is very sparse: $m+1 \in \{1,11,43,19,683,2731,...\}$ (for $k \in \{3,5,7,9,11,...\}$ and accordingly $n \in \{2,1,1,3,1,...\}$).

Consequence: not all even $m$ occur in that set, so only few $m$ satisfy the equation (1).

Note btw., that the occurence of $n$ shows an interesting pattern with increasing $k$, which can be better be described by noticing, that $2^k+1= {2^{2k}-1\over 2^k-1}$ and the pattern for occurence of $3^n$ in $2^k-1$ is perhaps more common knowledge (and from which it is clear that the set of solutions for $m+1$ is very sparse).