Is this argument valid for showing given integers $n>1$ and $2^m > 3^n$, $(2^m - 3^n) \nmid 3^n - 2^n$

It occurred to me that this might be straight forward to prove. I wanted to check if my reasoning was correct.

I will be glad if anyone points out a mistake or a simpler, more general way to make the same argument.

Let:

• $m > 1, n> 1$ be integers
• $2^m > 3^n$

Claim:

$(2^m - 3^n) \nmid (3^n - 2^n)$

Argument:

(1) $2^m - 3^n \ne 3^n - 2^n$ since:

• Assume $2^m - 3^n = 3^n - 2^n$
• $2^n(2^{m-n} + 1) = 2\cdot3^n$
• which is impossible for $n > 1$

(2) Assume that there exists an integer $a>1$ such that:

$$a(2^m - 3^n) = 3^n - 2^n$$

(3) It follows that:

$$a(2^n(2^{m-n} - 1)) + a2^n - a3^n = 3^n - 2^n$$

$$a(2^n(2^{m-n} - 1)) = (a+1)(3^n - 2^n)$$

(4) But then:

$$(2^n(2^{m-n} - 1)) = \frac{a+1}{a}(3^n - 2^n)$$

which is impossible

Edit:

I missed the case where $2^n | (a+1)$. If I can find an argument against this case, I will post it. Otherwise, my argument is incomplete.

If you narrow your question to Collatz cycles (which I suspect to be the case), that is a question I asked myself (see EDIT point here: Showing $3^iq-2^{i-p}\neq2^pq-1$, with $p:=\lceil i\,\log_23\rceil$, $q:=\left\{\frac{2^{i}\,3^{2^{p-i-2}}}{2^p3^i}\right\}$, $i>6$) in the case of a non-trivial "1-cycle" of the Collatz conjecture. Note that this adds an additional constraint on $m$.

To have a one cycle you start at odd $a = b2^n - 1$, you climb up to $b3^n - 1$ and than divide by $2^{m-n}$ to reach $a$ again:

$$b2^n - 1=\frac{b3^n - 1}{2^{m-n}}$$

In other words $$a=\frac{(a+1)\frac{3^n}{2^n}-1}{2^{m-n}}$$ or $$a(2^m-3^n)=3^n-2^n$$

In a cycle, $\frac{3^n}{2^m}<1$ and with the above formula we can write

$$0 < \frac{2^m}{3^n} - 1 = \frac{2^{m-n} -1}{b3^n} = \frac{2^{m-n} - 1}{2^{m-n}(b2^n - 1) + 1} < \frac {1}{2^n}$$

For an $x>1$ you have $0<\log x<x-1$ so you can write the above formula as

$$0 < m \log 2 - n \log 3 < \frac{2^m}{3^n} - 1 < \frac {1}{2^n}$$

But with the Rhin bound: $|m\log 2 - n\log 3| > 0.00218n^{-13.3}$ and some manual check (for $n<96$ as shown by Steiner in 1977), we can see this is not possible ($2^m-3^n\nmid3^n-2^n$ at least with the additional constraint on $m$)