We first note that $(0, 0, 0)$ is obviously a solution of the system of equations. We will now show that there are no others.
Let $x = \frac{p}{q}$ with relatively prime integer values of $p$ and $q$ and $q > 0$. We then have
$$y = \left( \left( \frac{p}{q} \right)^2 + 1 \right)^3 - 1 = \frac{(p^2 + q^2)^3 - q^6}{q^6} = \frac{p^6 + qQ}{q^6} = \frac{r}{q^6},$$
and this fraction cannot be simplified, since $p$ and $q$ are relatively prime. Further substitutions then yield $z = \frac{s}{q^{36}}$ and $x = \frac{t}{q^{216}}$, and since these fractions similarly cannot be simplified, $q^{216} = q = 1$ follows. We see that $x$ (and also $y$ and $z$) must be integers. For integer values not equal to $0$, we have $(x^2 + 1)^3 > x^2 + 1 \ge x + 1$, and since equality must hold if the three equations are multiplied, this yields a contradiction. We see that $(0, 0, 0)$ is indeed the only solution, as claimed. $\square$