Divisibility of $2^n - 1$ by $2^{m+n} - 3^m$. For what values of $m,n$ natural, do $2^n - 1$ is divisible by $2^{m+n} - 3^m$?
Thank you very much.
 A: The question is indeed equivalent to find solutions for an 1-cycle in the Collatz-problem. Let's first rewrite your divisibility - criterion as a cofactored expression
$$2^n-1 =  (2^{n+m}-3^m) \cdot h \qquad \qquad \text{for some integer $h \ge 1$ }\tag 1$$
From the study of cycles in the Collatz-problem there is the theorem:   
$ \qquad \qquad$   There are no solutions for (1) except $m=n=h= 1$ .
There is a short note of Ray Steiner where he paraphrases his disproof of the possibility of a nontrivial 1-cycle in the Collatz-problem. His formula is identical except the naming of variables. He wrote me a short informal note which I adapt for the use here:

Briefly, my 1977 proof runs as follows. I will just give the steps,
  not the details here.  
1) Any circuit for the 3x+1 problem
     corresponds to an integer solution $(k, l, h)$ of the Diophantine equation
     $ (2^{k+l} - 3^k)h = 2^l -1 \qquad \qquad (*)$              
2) To show that the only integer solution of (*) is $(1,1,1)$
         First, reduce this to a problem in linear forms in logarithms:
       $$0 \lt | l/k - \log_2 3/2 | \lt 1/(k \cdot \ln 2 \cdot (2^k-1)) $$      
3) This shows that if $k \gt 4$ then $l/k$ must be a convergent in the continued fraction expansion of $\log_2(3/2)$  .     
4) By using a lemma of Legendre, one can prove that a partial quotient of this CF must exceed $10^{4690}$         
5). Using Baker's, or Rhin's theorem one finds a reasonable upper bound for $k$ and the denominators of all convergents in this range are all smaller than $2500$.      

There is a more involved discussion of this and extension from 2- up to 68-cycles done by John Simons and Benne de Weger in the early 2000's. See the references which I've provided in the wikipedia-article on the Collatz-problem.
