Once again we are looking for positive, coprime solutions. This means x and z are odd, else all variables would be even.
Looking mod 4, y must be even, else 1+2 = 1.
Rewrite x2 + 2y2 as (x+yq)×(x-yq), where q is the square root of -2. As before, only complex primes divide x+yq, x+yq and x-yq are coprime, all primes in x+yq are squared, x+yq is the square of some u+vq, and the following equations hold.
u > 0, v > 0, u and v coprime, u is odd:
x ← u2-2v2
y ← 2uv
z ← u2+2v2
As before, u and v are coprime, but this time u is odd, and v can attain any parity. Both y and 2uv are even, so components correspond. Make u and v positive, which makes y positive. However, if x is negative, we can't simply swap u and v as we did before. That's all right; we'll allow x to be negative when u < sqrt(2)v. Different u v pairs in the first quadrant cannot be associates, and cannot produce the same triple, thanks to unique factorization. Once again the map is 1-1.
When u = 5 and v = 2, x = 17, y = 20, and z = 33.
When u = 7 and v = 3, a same parity solution, x = 31, y = 42, and z = 67.
When u = 3 and v = 7, x = -89, y = 42, and z = 107.