If $n = 0$, using the inequalities $x^2 \ge x$ and its analogues, we deduce that $x$, $y$, $z \in \{0, 1\}$.

If $n \ge 1$, then $2$ divides $x^2 + y^2 + z^2$, and hence, either the three numbers are even, or one is even and the others are odd. In the former case, if we take $x = 2x_1 + 1$, $y = 2y_1 + 1$, $z = 2z_1$, we get $4(x_1^2 + x_1 + y_1^2 + y_1 + z_1^2) + 2 = 4(x_1 + y_1 + z_1 + 1)$, which is a contradiction.
Let us consider the case when the numbers $x$, $y$, $z$ are even. For $x = 2x_1$, $y = 2y_1$, $z = 2z_1$, we get $x_1^2 + y_1^2 + z_1^2 = 2^{n-1}(x_1 + y_1 + z_1)$, and thus, if $n = 1$, then $x$, $y$, $z \in \{0, 2\}$.
For $n > 1$, using the same argument, we deduce that if $x = 2^n x_n$, $y = 2^n y_n$, $z = 2^n z_n$, then $x_n$, $y_n$, $z_n \in \mathbb{Z}$ and $x_n^2 + y_n^2 + z_n^2 = x_n + y_n + z_n$, whence $x_n$, $y_n$, $z_n \in \{0, 1\}$, and thus $x$, $y$, $z \in \{0, 2^n\}$.