How to find the integer solutions of $\frac{2^m-1}{2^{m+x}-3^x}=2a+1$?

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Is there a way to find all integer triplets of $(x, m, a)$ for the following equation. $$\frac{2^m-1}{2^{m+x}-3^x}=2a+1$$

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@Next did already link to the other Q&A, so it is no more need to discuss this further. But I've looked at this one time with a slightly different focus, and may be the reformulation looks interesting for you for further experimenting.

Let for convenience $2a+1 = k$ and let us express $3^x $ in terms of $2^m$ such that $ 3^x = n \cdot 2^m + r $ where $0<r<2^m$

Then your formula $$ { 2^m- 1 \over 2^m2^x - 3^x } = 2a +1 $$ changes to $${ 2^m- 1 \over 2^m2^x - (n2^m + r) } = k\\ { 2^m- 1 \over 2^m (2^x - n) - r } = k\\ 2^m- 1 = k(2^m (2^x - n) - r)\\ 2^m = k 2^m (2^x - n) - (kr -1)\\ 1 = k (2^x - n) - {kr -1 \over 2^m}\\ k (2^x - n) = {kr -1 \over 2^m}+1 \qquad \qquad \text{where } {kr -1 \over 2^m}+1\le k\\ $$ The last form of this equation has now an additional interesting property. The rhs can now be at most equal $k$ (because $r$ is smaller than $2^m$) so on the lhs the term $2^x-n$ is not allowed to become greater than 1; thus so we need to have $n=2^x-1$. But if we look now at the decomposition of $3^x$ then we see, that we must have that $3^x = n \cdot 2^m +r = (2^x-1) \cdot 2^m + r = 2^{x+m} - 2^m+r $and the difference between the perfect power of 2 and that of the perfect power of 3 is expected to be $2^{x+m}-3^x = 2^{x+m} - (2^{x+m}-2^m+r) = 2^m-r$ . But this does happen only in the "trivial" small case(s).
The relation of neighboured perfect powers of 2 and 3 have been much studied, and perhaps it is also interesting for you to look at the "Waring's" problem to see some more general relations.

One more tiny remark: we have not only a focus on the difference between perfect powers here, but also some modularity condition: the value $2a+1 = k$ must be the modular inverse of the residual $r = 3^x - n \cdot 2^m$ and is thus restricted by this rule ... and thus one might look at it with even a bit more couriosity...