$3^x-2^y=17$ where $x$ and $y$ are both positive integers First, I was wondering what would the solution(s) be to the equation $3^x-2^y=17$ such that $x$ and $y$ are both positive integers. I couldn't find any small possibilities. Is there a proof that such a solution exists? How about the general case $3^x-2^y=z$? Is there exactly one x and y positive integer solution for every positive integer z (not divisible by 2 or 3 that is)? Thanks in advance!

So the constant is $17$ instead of $11$. Well, working in the field $\mathbb F_{17}$ one has the equation $$3^x=2^y$$ Noticing now that in this field, $\color{green}{2=3^{14}}$ and $\color{green}{3^{16k}=1}$ it follows the equivalent equation  (in $\mathbb F_{17}$)
$$3^x=3^{14y} \equiv 3^{x+16k}=3^{14y}$$ It follows (in $\mathbb Z$ now) $$x+16k=14y \quad (*)$$ The equation $(*)$ has an infinity of integer solutions $(x,y,k)$ nevertheless it is clear that not all of them fit in the equation $3^x-2^y=17$.
We use the theorem of Stroeker and Tijdeman quoted by @Andre Nicolas according to which if we find a solution or we prove there is none,  we finished (my will has weakened with 11 previously put in place of 17). 
We find easily $(x,y,k)=(4,6,5)$ as a fitting solution. Thus $$\color{red}{3^4-2^6=17}$$ shows the only solution.
