With the following equation-
$\frac{(3^x-2^x)}{(2^{(x+y)}-3^x)} = z $
What $x,y,z$ (all integers) satisfy this?
The trivial solutions are-
$X=1,y=1,z=1 ,X=2,y=1,z=-5 ,X=1,y=0,z=-1 $
This is my own “restatement” of the impossibility of the $k=1$ cycle for collatz sequences and it would be incredibly helpful if someone could prove it in this form (it should be possible to prove that no integer solutions for $x,y,z$ exists other than the trivials)
A "1-cycle" in Collaz implies
$$\frac{2^m}{3^n} - 1 < \frac {1}{2^n}$$ $$2^m-3^n < (\frac {3}{2})^n$$ $$2^m-3^n < 1.5^n$$
but from this we know that ($n>5$)
$$2^m-3^n \ge 2^{\lceil n \cdot log_2(3)\rceil}-3^n> 2^n$$ $$2^m-3^n > 2^n>1.5^n \implies \text{contradiction}$$
So for the value of $m=x+y$ and $n=x$ that you consider (Collatz cycle where $m$ is believed to be equal to $\lceil n\cdot log_2(3)\rceil$), and outside the exception for $n=1$, we have
$$2^m - 3^n \nmid 3^n - 2^n$$
EDIT
Another approach for the first part in the context of a 1-cycle ($a,b$ from the link): $$\frac{3^n-2^n}{2^m-3^n}=a=b\cdot2^n-1\ge2^n-1$$ $$\implies 2^m-3^n \le \frac{3^n-2^n}{2^n-1}<\frac{3^n}{2^n}$$ $$\implies 2^m-3^n < 1.5^n$$
For negative values, use the same reasoning (but here $2^m<3^n$):
$$\frac{3^n-2^n}{2^m-3^n}=a=-b\cdot2^n-1\le -2^n-1$$ $$\implies 2^m-3^n \ge \frac{3^n-2^n}{-2^n-1}$$ $$\implies 3^n-2^m \le \frac{3^n-2^n}{2^n+1}<\frac{3^n}{2^n}$$ $$\implies 3^n-2^m < 1.5^n$$
but from the same link above ($n>2$ and $m$ is smaller than $\lceil n\cdot log_2(3)\rceil$)
$$2^{m+1}-3^n\le2^{\lceil n\cdot log_2(3)\rceil}-3^n<3^n-2^n$$ $$2\cdot 2^m-2\cdot 3^n<-2^n$$ $$3^n-2^m>2^{n-1}>1.5^n \implies \text{contradiction}$$
with the exception you found for $n\le2$