For the equation:
I want to prove it has no solutions with x,y being positive integers.
Is there a general method for solving this type of equation? (It looks vaguely like a Pell equation, but not close enough that I can see how to solve it with standard methods)
If not, is there an elegant method to prove it for the particular case here (with $A=12$ and $B=5$)?
What I've tried
Thinking modulo 12, the LHS = 0, and the RHS is 0 if and only if y is even. Writing $y=2z$, I can then factorize the RHS into $(5^z+1)(5^z-1)$
Both factors are even and by thinking modulo 3, only one of these factors can be divisible by 3. So I conclude that I need something like $5^z+1=2^?3^x$ and $5^z-1=2^?$ or vice versa.
Subtracting these equations I need $2=2^?3^x-2^?$.
If I now think in binary, these equations look like $10_2 = (11_2)^x100..00_2 - 100...00_2$.
It seems to make sense (but I don't see how to mathematically express this idea) that the only way this equation will work is as $5^1+1=2.3$ and $5^1-1=2.2$ but this solution results in a LHS of 24, which is not a power of 12.
However, I feel there must be a less convoluted proof!