$3^x-2^y=17$ 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!

Because we know, that there is one solution possible
$$  3^4-2^6 = 17 \\$$ we rewrite the problem a bit:
$$ 3^{x+4} - 2^{y+6} \overset?= 17 = 3^4-2^6  \qquad \qquad \text{with } x,y \gt 0 \tag 1$$
and reformulate for readability
$$ {3^{x} - 1 \over 2^6} \overset?={2^{y}-1  \over3^4 } \qquad \qquad \text{with } x,y \gt 0\tag 2
$$
First, to get the numerators containing the denominators as factors their exponents must have the following structure:
$$
 {3^{2^4 x_1} - 1 \over 2^6} ={2^{2 \cdot 3^3 \cdot y_1}-1  \over3^4 }
\qquad \qquad \text{ where } 2 \not \mid x_1 \text{ and } 3 \not \mid y_1 \tag 3
$$

Now let's assume $x_1 = y_1 = 1$. We get the primefactors for the lhs as
$$  {3^{2^4 x_1} - 1 \over 2^6} = 5.17.41.193 \tag{ lhs} $$
and for the rhs as 
$${2^{2 \cdot 3^3 \cdot y_1}-1  \over3^4 } = 7.19.73.\text{<big>} \tag { rhs} $$ 
So far the lhs and rhs have no common factor and thus can trivially not be equal.               

We can now try iteratively to adapt the exponents such that either the both sides have at some iteration the same primefactorization and are thus equal or that we arrive at an impossibility and thus at a contradiction.     
For instance, to make the rhs containing the primefactor $193$ the exponent must contain the order of $193$ to base $2$, such that $193 \mid \text{ <rhs> }$ . This requires that in the exponent we have the value $96 = 2^5 \cdot3$    and we can write
$${2^{2^5 \cdot 3^3 \cdot y_2}-1  \over 3^4 } = 193.\quad 5.7.13.17.19.37.73.97.109.241.257.\text{<big>} \tag { rhs} $$ 
and so indeed the rhs contains now the primefactor $193$. But this introduced a lot of additional primefactors, which must now occur in the lhs, too.     
Let's look at the primefactor $257$. To get it in the lhs we need in the exponent of the lhs the value $2^8$. But this contradicts the restriction in $(3)$ where we find that $2 \not \mid x_1$ and we need now $2^4 \mid x_1$ to allow the primefactor $257$ to appear.         
Thus we have a contradiction, and we can state:   
Theorem: There is no second solution besides $x=y=0$ in $(3)$ and thus not in (1)
