If $n\ge 2$, then $x^2+y^2+z^2\equiv 0\pmod{4}$. Squares are $0$ or $1$ mod $4$ because, e.g., $(2h+1)^2=4(h^2+h)+1$, $(2h)^2=4h^2$, $h\in\mathbb Z$. We get $x=2x_1$, $y=2y_1$, $z=2z_1$, $x_1,y_1,z_1\in\mathbb Z$. $x_1^2+y_1^2+z_1^2=2^{n-1}(x_1+y_1+z_1)$. If $n-1\ge 2$, then $x_1=2x_2$, etc., $x_2^2+y_2^2+z_2^2=2^{n-2}(x_2+y_2+z_2)$, ... Continue this until you get $x_k^2+y_k^2+z_k^2=2(x_k+y_k+z_k)$. Then $x_k^2+y_k^2+z_k^2\equiv 0\pmod{2}$, also $a^2\equiv a\pmod{2}$ for all $a\in\mathbb Z$, so $x_k^2\equiv x_k\pmod{2}$, etc., so $x_k+y_k+z_k\equiv 0\pmod{2}$, so $4\mid 2(x_k+y_k+z_k)$, so $x_k^2+y_k^2+z_k^2\equiv 0\pmod{4}$, so $x_k=2x_{k+1}$, $y_k=2y_{k+1}$, $z_k=2z_{k+1}$, so $x_{k+1}^2+y_{k+1}^2+z_{k+1}^2=x_{k+1}+y_{k+1}+z_{k+1}$. We have $x_{k+1}^2\ge x_{k+1}$, $y_{k+1}^2\ge y_{k+1}$, $z_{k+1}^2\ge z_{k+1}$, so $x_{k+1}^2+y_{k+1}^2+z_{k+1}^2\ge x_{k+1}+y_{k+1}+z_{k+1}$ and equality holds if and only if $x_{k+1}\in\{0,1\}$ and $y_{k+1}\in\{0,1\}$ and $z_{k+1}\in\{0,1\}$. If $n=1$ or $n=0$, then it's similar. We get that $\exists$ $m\ge 0$, $m\in\mathbb Z$ such that $x\in\{0,2^m\}$, $y\in\{0,2^m\}$, $z\in\{0,2^m\}$. Checking cases $x=y=z=2^m$, $x=0$, $y=z=2^m$, $x=y=0$, $z=2^m$, $x=y=z=0$ gives that $(x,y,z)=(0,0,0)$ is a solution and if $(x,y,z)\neq (0,0,0)$, then $m=n$. All the solutions are $(x,y,z)=(0,0,0)$, $(0,0,2^n)$, $(0,2^n,2^n)$, $(2^n,2^n,2^n)$ for all $n\in\mathbb N$ and permutations.

summary of post #2: $2^n|x,y,z$ dividing by $2^{2n}$ , we get $x_1^2+y_1^2+z_1^2=x_1+y_1+z_!$ where each variable is 0 or 1.