Consider $mod\ 7$. We know that $a^2\equiv0,1,2,4(mod\ 7)$ for integer $a$ and because $7$ divides $2016$ and $77$, the only possibilities are if both $x$ and $y$ are divisible by $7$. Write $x=7x_1$ and $y=7y_1$. Plugging to original equation we get that $49({x_1}^2+{y_1}^2)=3\cdot2016^z+77$. Now consider 2 cases: $1)$ $z\ge 2$ We see now that $49\mid2016^z$ and $49\mid LHS$ so because $49\nmid77$ we get contradiction. $2)$ $z\le 1$ Doing casework on $z=1$ and $z=2$ gives 3 solutions: $(4,8), (14,77), (35,70)$.

uh... i think if $z \le 1$, $(x_{1}^{2}+y_{1}^{2})=125$ and maybe the answer is $ (14, 77), (35, 70)$... isn't it?

Just work (mod 7)

(x,y,z) = (4,8,0) its non-negative, z can be 0

Clearly $x,y\neq0$, because when either $x\text{ or }y=0$ we obtain $3\cdot2016^z+77\equiv2\pmod 3\text{ which forces }x^2,y^2\equiv 2\pmod 3$ which is clearly a contradiction. Therefore $x,y\in\mathbb{Z}^+$ If $z=0$: We obtain the the equation $x^2+y^2=80$ which implies that $x,y\le8$, moreover individually checking the cases one by one we obtain $(x,y)=(4,8)$ to be the only valid solution. Thus $(x,y,z)=(4,8,0)$ is one solution, Therefore from now on assume that $z\in\mathbb{Z}^+$ Now notice that $3\cdot2016^z+77\equiv 0\pmod 7\text{ which forces }x^2+y^2\equiv0\pmod 7$. However notice that $n^2\equiv 1,2,4\text{ or }0\pmod 7$ thus for $x^2+y^2\equiv0\pmod 7$, $x^2\text{ and }y^2$ must both be $\equiv0\pmod 7$ which is only the case when both $x\text{ and }y$ are divisible by $7$ So let $x=7n\text{ and }y=7k$ Thus the original equation transforms into $49n^2+49k^2=3\cdot2016^z+77$ Moreover notice that for $z\ge2$ we obtain $3\cdot2016^2+77\equiv28\pmod {49}\text{ while }49n^2+49k^2\equiv0\pmod {49}$ which is a contradiction. Therefore $z=1$ Thus the equation becomes $49n^2+49k^2=6125\Longrightarrow n^2+k^2=125\therefore n,k\le11$ Checking these cases we obtain that $(n,k)=(2,11)\text{ and }(5,10)$ to be the only solutions, this furthermore implies that $(x,y)=(14,77)\text{ and }(35,70)$ Therefore $(x,y,z)=(14,77,1)\text{ and }(35,70,1)$ are the other two solutions to the equation. So, in conclusion $\boxed{(x,y,z)=(4,8,0),(14,77,1)\text{ and }(35,70,1)\text{ are the only solutions for }x,y,z\in\mathbb{Z}_{\ge0}}$ $\blacksquare$.