Diophantine equation $3^{m}x+3^{m}-3^{m-i} 2^{i} +3^{m-i} 2^{i+s} -2^{m+s}=x 2^{m+s}$

Question

As part of my research and my calculations I got the following diophantine equation. I do not have much experience with the diophantine equation. Is there any known method to solve it? Any help is appreciated:

$$3^{m}x+3^{m}-3^{m-i} 2^{i} +3^{m-i} 2^{i+s} -2^{m+s}=x 2^{m+s}$$

with

$m,i,s,x\in\mathbb{N}$

Thanks.

Answer 1

I've answered the question $$ {2^s-1\over 2^{m+s}-3^m}$$ (as formulated in @mark's comment) in the answer where @barto has linked to. There is a proof of Ray Steiner (1977) that

the only solution is $(m,s)=(1,1)$.

(See details in the linked other question)