I am attempting to generalize an answer given by Will Jagy here.
I may have made a mistake since the conclusion is stronger than I typically have seen and my argument is pretty much the same as Will's. If that is the case, then I would appreciate if someone could call out my mistake.
Let:
- $m>2, n>1, x>0, y>0, c>0$ be integers with:
$$2^m - 3^n = c$$
Does it now follow that:
$$2^{m+x} - 3^{n+y} \ne c$$
Here's my thinking:
(1) Assume that $2^{m+x} - 3^{n+y} = 2^{m} - 3^n = c$
(2) $2^m\left(2^x - 1\right) = 3^n\left(3^y - 1\right)$
(3) $3^y - 1 \equiv 0 \pmod {2^m}$
(4) From Euler's Theorem, $2^{m-1} | y$ since from Euler's Totient Function $\varphi(2^m) = 2^{m}\left(1 - \dfrac{1}{2}\right) = 2^{m-1}$
Note: I believe that Carmichael's Theorem can be used to establish that $\varphi(2^m) | y$ since $3$ is an odd prime.
(5) So there exists an integer $a > 0$ with:
$$2^m\left(2^x - 1\right) = 3^n\left(3^{2^{m-1}a} - 1\right) = 3^n(3^{2^0a} - 1)(3^{2^0a} + 1)(3^{2^1a}+1)(3^{2^2a}+1)\dots(3^{2^{m-2}a}+1)$$
(6) For each $0 \le i \le m-2$, $2 | (3^{2^ia} \pm 1)$ so $2^m | (3^y - 1)$
(7) Regardless of whether $a$ is even or odd, $8 | (3^a - 1)(3^a + 1)$
(8) But then $2^{m+1} | (3^y - 1)$ which is impossible since $2^{m+1} \nmid 2^m(2^{x} - 1)$
Edit: Added note about Carmichael's Theorem based on comment.
Edit 2: I applied Carmichael's Theorem incorrectly. For $2^{m}$, Carmichael's Theorem says $\dfrac{1}{2}\varphi(2^n)$
For example:
$$8(2^2 - 1) = 3(3^2 -1 )$$
In this case, $\varphi(8)=4 \nmid 2$ but $\dfrac{1}{2}\varphi(8) = 2 | 2$
This mistake invalidates the argument.