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