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This question arose as an attempt to answer the following question Relaxed Collatz 3x+1 conjecture. I wanted to show that there is a solution of the equation $2^{k}=3^{z}(2n+1)-1$ for each $n\geq 2$, where $k,z,n\in\mathbb{N}$. But even a special case has put me in a dead end.

Are there infinitely many solutions of the equation $2^k=3^z-1$, when $z\rightarrow \infty$? First solution: $2^3 =3^2 -1$.

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  • $\begingroup$ math.stackexchange.com/questions/958304/… $\endgroup$
    – j.c.
    Commented Sep 16, 2017 at 15:48
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    $\begingroup$ This question is answered by the answers to Are there any solutions to $2^n-3^m=1$ as well as mathoverflow.net/a/29928 $\endgroup$
    – j.c.
    Commented Sep 16, 2017 at 15:54
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    $\begingroup$ More generally, for fixed $a,b\in\mathbb Z$ with $|a|\ge2$ and $|b|\ge2$, there are only finitely many $k,z\in\mathbb N$ such that $$ a^k = b^z-1.$$ I understand why people closed this question, since the result is well known. On the other hand, the solution requires non-trivial results from Diophantine approximation (at least Thue 1909 result), and effective estimates for the problem may need linear forms in logs, so one could make the case that this question is appropriate for MO. $\endgroup$
    – Joe Silverman
    Commented Sep 16, 2017 at 16:09
  • $\begingroup$ The question has two parts; only the second one (the special case) is a duplicate. The first part can be rewritten as $$ { 2^k+1\over w } = 3^z $$ with $w \ge 5 $ is odd. Of course, for every $k$ we find some $w$ giving a solution for $z$ (following from "Little Fermat" from where $k$ is a relatively simple function of $z$). However, whether every w can occur I don't know at the top of my head, but I think that actually many $w$ don't occur. $\endgroup$
    – Gottfried Helms
    Commented Nov 25, 2017 at 11:49

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The answer is no. Actually Catalan's conjecture, or Mihăilescu's theorem, suggests that the only solution of your equation $2^k = 3^z - 1$ is $k=3,z=2$.

As the comment below has mentioned, I missed another solution $k=z=1$.

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    $\begingroup$ Not "suggests" but "shows". $\endgroup$
    – GH from MO
    Commented Sep 16, 2017 at 18:28
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    $\begingroup$ Don't forget about $k=z=1$ (as an answer to the original question). $\endgroup$
    – Pace Nielsen
    Commented Sep 16, 2017 at 20:59