Solving for integers $m>8, n > 0$ for $2^m - 3^n = 13$ Let $m,n$ be integers such that:
$$2^m - 3^n = 13$$
$m > 8$ since $2^8 - 3^5 = 13$.
I am trying to either find a solution or prove that no solution exists.
I tried to use an argument similar to this one for $2^m - 3^n = 5$ where $m > 5$.
$3^n \equiv -13 \pmod {512}$ if and only if $n \equiv{69} \pmod {128}$
Is there a straight forward way to complete the argument?  Is there a better way to answer the question for $2^m - 3^n = 13$ with $m > 8$?

Edit:
Adding context for question:
I have been trying to understand why it is so difficult to establish a lower bound to $2^m - 3^n$ when $2^m > 3^n$.  This came out of thinking about the Collatz Conjecture.
It occurs to me that for $m \ge 3$, $2^m - 3^n$ is congruent to either $5$ or $7$ modulo $8$  (since $-3^{2i} \equiv 7 \pmod 8$ and $-3^{2i+1} \equiv 5 \pmod 8$)
For $2^m - 3^n \equiv 7 \pmod {12}$, $m$ and $n$ are even, so the lower bound is at least $3^{n/2}$ since:
$$2^m - 3^n = (2^{m/2} - 3^{n/2})(2^{m/2} + 3^{m/2}) > 0$$
and
$$2^{m/2} - 3^{m/2} \ge 1$$
So, to reach a lower bound, I need to better understand the implications of $2^m - 3^n \equiv 5 \pmod 8$ and $2^m - 3^n \equiv 7 \pmod 8$ when $2^m - 3^n \not\equiv 7 \pmod {12}$.
I am also working to better understand this blog post by Terence Tao.

Edit 2:
I think that I can complete the argument using the congruence classes of $257$.
Here's my thinking:
Since $n \equiv 69 \pmod {128}$, $3^n \equiv 224$ or $33 \pmod {257}$
Then $2^m \equiv 3^n + 13 \pmod {257}$ which means $2^m \equiv 237$ or $46 \pmod {257}$
But there is no such solution of $2^m$ since for each $2^m$, there exists an integer $i$ such that $2^m \equiv \pm 2^{i} \pmod {257}$ and there is no such $i$ where $\pm 2^{i} \equiv {237} \pmod  {257}$ or $\pm 2^{i} \equiv {46} \pmod {257}$

we have $$ 256(2^x - 1)  = 243(3^y - 1) $$
and we assume $x,y \geq 1$
since $2^x \equiv 1 \pmod {243}$ we calculate that $162 | x.$
Next, $2^{162} - 1$ is divisible by the prime $262657.$
since $3^y \equiv 1 \pmod {262657}$ we calculate that $14592  | y.$  In particular, $256|y.$
Well, $3^{256} - 1$ is divisible by 1024. So that
$ 256(2^x - 1)  $ is divisible by 1024, which is a contradiction of $x \geq 1$
similar:
https://math.stackexchange.com/users/292972/gyumin-roh
Exponential Diophantine equation $7^y + 2 = 3^x$   Gyumin Roh answer from 2015
Elementary solution of exponential Diophantine equation $2^x - 3^y = 7$.
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100
http://math.stackexchange.com/questions/2100780/is-2m-1-ever-a-power-of-3-for-m-3/2100847#2100847
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