<s>[INST] Divisibility problem: $ \frac{3^{m}}{2^{n} - 3^r} $ 
Is divisible a power of 3 for a difference of powers of 2 and 3? That
  is, can result, this division, in an integer? $$ \frac{3^{m}}{2^{n} -
3^r} $$ where $n,m,r$ natural number.

Edit: $n>r$, $r=m+1$.
 [/INST] HINT:
We need $\displaystyle 2^n-3^r=\pm3^s$ where $s\le m$
$\displaystyle 2^n=3^r\pm3^s$
Case $1:$ Taking the $+$ sign, $2^n=3^r+3^s=3^{r-s}(3^s+1)$ (WLOG $r\ge s$)
As $(2,3)=1,$ we need $3^{r-s}=1\iff r=s$ and $2^n=3^r+1$
Case $2:$ Taking the $-$ sign, $2^n=3^r-3^s=3^{r-s}(3^s-1)$ (we need $r\ge s$)
Like the previous case, we must have $r=s\implies 2^n=3^r-1$
See Catalan's Conjecture
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