We're given a*(x + y + 1) = 2xy and b*(x + y - 1) = x^2 + y^2 - 1.

Add the two equations and factor.

a*(x + y + 1) + b*(x + y - 1) = (x + y + 1)*(x + y - 1)

RHS and first factor of LHS are divisible by (x + y + 1), so b*(x + y - 1) = x^2 + y^2 - 1 is factored by both (x + y - 1) and (x + y + 1).

But the product of those factors is already greater than x^2 + y^2 - 1 and will therefore be equal to an integer multiple of that number.

Write (x + y - 1)*(x + y + 1) = k*(x^2 + y^2 - 1), consider the result as a quadratic in x, and look at the discriminant. Then k=2 gives x = (y + 1) or x = (y - 1), matching the results of the first two posts, but higher values of k require that (-p)*y^2 + (p + 1) be a square, thus y = 1, but with x a fraction.