I understand your problem well. You can find an expression for x,y and k.
Similarly,
Thank you very much for the hint. Could you please take a quick look at what I came up with as I was waiting for your reply?
If y=0, we have , so and are powers of 2, and the only powers of 2 that differ by 2 are 2 and 4, so (x,y,k)=(3,0,3).
If y>0, is a quadratic residue modulo 5, so x must be even. Hence . The factors are powers of 5, say and .
We now have . Since the right side is not divisible by 5, we must have n=0(is it correct?). Then, for m=1, we get x=2, and so (x,y,k)=(2,1,3). But what next? How do I prove that there are no more solutions?