Are you smarter than me? 41

First of all, note that $2016 \equiv 0 \pmod{4}$ implies that none of $x,y,z$ is odd.

Because if any out of $x,y,z$ is odd, $x^2,y^2,z^2$ will correspondingly be congruent to $1$ $\pmod 4$ , and no combination of 3 or less 1s will give a number divisible by 4.

Hence we conclude $2 \mid x , 2\mid b , 2\mid c$

---

We can say $a=\frac{x}{2} , b = \frac{y}{2} , c = \frac{z}{2}$ , and number of solutions of $a^2+b^2+c^2 = \frac{2016}{4}$ will be same as of original equation.

$a^2+b^2+c^2 = 504$ .

Now because $4 \mid 504$ , we conclude that again all of $a,b,c$ are even.

---

So now problem reduces to integer solutions of $j^2+k^2+l^2 = 126$

As $4 \nmid 126$ , we conclude that $\textbf{exactly}$ 2 out of $j,k,l$ are odd . And, the original numbers $x,y,z$ are basically $4j,4k,4l$ .

---

For simplicity, first let's find positive integer solutions such that $a\leq b\leq c$ .

Now by observation, we have only the following solutions

$(1,2,11) \\ (1,5,10) \\ (3,6,9)$

For ordered triples, there will be $3\times 3!$ of these solutions.

Now as we want integer solutions, negatives would work too! Each of the 3 can be either +ve or -ve, giving $2\times 2\times 2 = 8$ combinations of +ve and -ve.

Hence answer is $3\times 3! \times 8 = 3\times 6\times 8 = \boxed{144}$ .