Complete The Last Square

We multiply by $2$ to get $(x+y)^4 + (x-y)^4 = (2z)^4.$ By Fermat's Last Theorem we know that a solution must have one of the terms equal to 0. Since these are non-negative integers

Case 1: $x + y = 0$
Then $x = y = 0$ . Clearly $z = 0$ .

Case 2: $x -y = 0$
So $x = y$ and $x+y = 2z$ . Thus $x = y = z$ .

There are $\boxed{2017}$ solutions, namely $(0,0,0), (1,1,1), \ldots, (2016,2016,2016)$ .