Integer solution of equation I was thinking about different types of equations in two variables. My motivation for this is purely intrinsic. I just wanted to know what we can say about analytic methods of solving different equations. On the course of this, I came across the following:

$5^x - 3^y = 2$

My guess is that the only solution in the positive integers is $x=y=1$. Also Wolfram Alpha is telling me that. Is there a way of approaching this problem.
Of course we can write $y = \frac{\log(5^x - 2)}{\log(3)}$, but I am not sure that this helps.
Thank you very much for your help!
 A: Firstly, we start with some modulus arguments. If we take the whole sequence modulo $3$, then we have that $5^x\equiv2$. In particular, $x\equiv1\pmod2$. Similarly, taking the sequence modulo $5$ tells us that $y\equiv1\pmod2$ as well.
The next step is to write $5^{x+1}=p^2$ and $3^{y-1}=q^2$. The equation then becomes, after multiplication by $5$, $$p^2-15q^2=10.$$ This is a generalised pell equation, for which the solutions are given here (Theorem 3.3). In particular, we solve the pell equation $$a^2-db^2=1$$ for $d=15$, which yields a primitive solution $(4,1)$, and then find all solutions to $$x^2-dy^2=n$$ for $n=10$ such that $|x|\le \sqrt{10(4+\sqrt{15})}\approx8.9$.
The only such solution, as you may have guessed, is $(5,1)$.
Finally, we compose the two solutions to find all solutions to the original equation, $$p^2-15q^2=10.$$ This is a simple recurrence, beginning with initial term $p_0=5,q_0=1$ with the conditions $$p_{n+1} = 4p_n + 15q_n,\\q_{n+1}=p_n+4q_n.$$
The trick to notice here is that this sequence can hardly ever have a power of $5$ as $p_n$ (which was, after all, the original question!) so we restrict ourselves to looking at powers of $5$. A straightforward induction tells us that $p_n$ satisfies $p_0=5,p_1=35,$ and $$p_{n+2}=8p_{n+1}-p_{n}$$ which is handy.
If you think about the condition that we don't want any other factors other than $5$ to appear, this sounds awfully reminiscent of Zsigmondy's Theorem -- indeed, the generalisation to Lucas sequences instantly reduces it to the finite case of $n\le30$.
>>> def is_power_of_5(n):
        while n%5==0:
            n//=5
        return n==1

>>> arr = [5,35]

>>> for i in range(40):
        arr.append(arr[-1]*8-arr[-2])

>>> for i in arr:
        if is_power_of_5(i):
            print(i)    
5

Hence $(1,1)$ is indeed the only solution to the question.
