Solving for y gives
y = 1 + 4x/(x^2 - 2x -1)
X is relatively prime to x^2 - 2x -1, so either
(a) x = 0 or
(b) 4 is divided by x^2 - 2x -1
In the 2nd case, x^2 - 2x - 1 = p where p is in (1, 2, 4, -1, -2, -4)
Solving that quadratic equation gives
x = 1 +/- sqrt(p+2)
if p = -1, then x is 0 or 2
if p = -2 then x is 1
if p = 2 then x is -1 or 3
So, the only solutions are:
(0,1), (2,-7), (1,-1), (-1,-1) and (3,7)