[Let (n,x,y,z) ∈ \mathbb {Z}.
Given x2 + y2 = n + z2, let x = z. Thus, we have y2 = n. Since y ∈ \mathbb {Z}, y2 is an integer as well by the closure properties of the integers, and n is an integer. Now, for all y, n = y2, and because there are infinite integers y, there are infinite integers y2, and thus infinite solutions exist where x = z and n = y2.
Now, for if you meant distinct solutions where x ≠ y ≠ z ≠ n, take any Pythagorean triple (x,y,z). Replace z2 with z = m - 1. z2 = (m-1)2 = m2 - 2m + 1. The equality x2 + y2 = m2 -2m +1 still holds if you let n = 2m - 1. However, will n be distinct from x or y now that n = 2m - 1? Because this is a Pythagorean triple, we know x < m - 1 and y < m - 1, and thus x or y couldn't possibly be 2m - 1 as 2m - 1 > m - 1 = z for m > 0. Finally, since there are infinitely many Pythagorean triples, there are infinitely many solutions to x2 + y2 = n+ z2 given that you let z = m - 1 and n = 2m - 1. All algebraic manipulations are justified under the the closure properties of the natural numbers, and the fact that there are infinitely many Pythagorean triples as large as one would desire.]