The equation can be written equivalently as $(2x - 1)^2 + (2y - 1)^2 + (2z - 1)^2 = 7$.

If $(x, y, z)$ would be a solution of this equation with rational components, denoting $2x - 1 = \frac{a_1}{b_1}$, $2y - 1 = \frac{a_2}{b_2}$, $2z - 1 = \frac{a_3}{b_3}$, one would have integers $a_1, b_1, a_2, b_2, a_3, b_3$ satisfying the equality $(a_1 b_2 b_3)^2 + (b_1 a_2 b_3)^2 + (b_1 b_2 a_3)^2 = 7 (b_1 b_2 b_3)^2$.

This would lead to the existence of four integers $a, b, c, d$ such that $a^2 + b^2 + c^2 = 7d^2$. If the greatest common divisor of $a, b, c, d$ is $k$, dividing by $k^2$ one would obtain a solution $(a, b, c, d) \in \mathbb{Z}^4$ of the equation $a^2 + b^2 + c^2 = 7d^2$ with $\gcd(a, b, c, d) = 1$. Then $a, b, c, d$ cannot all be even.

But a perfect square is congruent to either $0$, $1$, or $4$ modulo $8$, so the left hand side can only be $1$, $2$, $3$, $5$ or $6$ modulo $8$, while the right hand side, $7d^2$, is $0$, $4$ or $7$ modulo $8$. In conclusion, the equality cannot hold.