Take the equation mod 4, then we have {0 or 2} + 1 = {0 or 1} mod 4. This makes sense only when the equation is 0 + 1 = 1. Thus x is even and y is odd.
Let x=2z. Then (2^z)^2 + 17 = (y^2)^2. Move the 2^z term to the right and apply a difference of squares to yield 17 = (y^2 - 2^z) * (y^2 + 2^z).
17 has two integer factorizations: 1*17 and -1*-17. y^2 and 2^zmust be positive thus the only useable factorization is 1*17. Then y^2 + 2^z = 17 and y^2 - 2^z = 1.
Solving for y^2 and 2^z gives y^2=9 and 2^z = 8. Then y=3, z=3, and x=6. Then there is one answer: (x,y) = (6,3).
Solution
