ok I think i got it!

 

Clearly (x,y) = (0,±2) gives one pair of solutions, and there are no solutions for x=1. Equally clearly there can be no solutions with x<0.

If x>1, let . Then ,

or .

 

Therefore . So one of the numbers y+1+w, y-1-w must be a power of 2, and the other must be 7 times a power of 2.

Suppose that . Then . Thus . The left side of this equation is an odd multiple of 2, hence so is the right side, and this can only happen if n=1. Then , from which m=5 and x=4. This gives the solutions (x,y)=(4,±23).

The other possibility is that . A similar analysis to the previous paragraph leads to the equation . But a power of 2 can only be congruent to 1, 2 or 4 (mod 7), not to -1. So this equation has no integer solutions.

Therefore the only solutions are (x,y) = (0,±2) or (4,±23)!