Solution:

The only solution is $(6,10,15)$. First observe that $45$ divides $z^{3}$, and hence $z$ is divisible by $3 \cdot 5=15$. Similarly $8$ divides $x y^{2}$, and therefore $x$ and $y$ cannot both be odd. Set $z=15 w$ and distinguish 3 cases:

(1) $x$ odd: then $8$ divides $y^{2}$, so $4$ divides $y$ and, setting $y=4 v$, the system becomes
$$\left\{\begin{array}{l} 2 x v^{2}=75 w^{3} \\ x v w<\frac{50}{3} \end{array} ight. \right.$$
impossible because from the first equation it follows that $15$ divides $x v$ and $2$ divides $w$, whence $x v w \geq 30$.

(2) $y$ odd: then $8$ divides $x$ and, setting $x=8 u$, the system becomes
$$\left\{\begin{array}{l} u y^{2}=75 w^{3} \\ u y w<\frac{25}{3} \end{array} ight. \right.$$
which is again impossible because $15$ divides $u y$, so $u y w \geq 15$.

(3) $x$ and $y$ even: setting $x=2 u, y=2 v$ the system becomes
$$\left\{\begin{array}{l} u v^{2}=75 w^{3} \\ u v w<\frac{50}{3} \end{array} ight. \right.$$
As before we have that $15$ divides $u v$, so the second inequality implies $u v=15, w=1$ and, from the first, $u v^{2}=75$; hence $v=5, u=3, w=1$.

The only solution is $(6,10,15)$.

First we note that both $3$ and $5$ divide $8 z^{3}$, so, since $8$ is coprime to both, $3$ and $5$ must divide $z^{3}$ and hence also $z$. Let us then set $z=15 z'$, so that the first equation becomes $x y^{2}=600 z'^{3}$ with the condition $x y z'<1000 / 15<67$. If $z'$ were greater than or equal to $2$ we would have
$$x^{2} y^{2} \geq x y^{2}=600 z'^{3} \geq 600 \cdot 2^{3}=4800 .$$
This would imply $x y>69$, in contradiction with the condition $x y z'<67$. Therefore $z'=1$ and the equation becomes $x y^{2}=600$ with the condition $x y<67$. If $x$ were greater than or equal to $8$ we would have
$$x^{2} y^{2} \geq 8 x y^{2}=8 \cdot 600=4800 .$$
This would again imply $x y>69$, giving another contradiction, so we have $x<8$. Since $y^{2}=\frac{600}{x}$, we have that $\frac{600}{x}$ is the square of an integer. An easy check shows that the only possibility is $x=6$, from which we obtain $y=10$ and $z=15$. On the other hand, the triple $(6,10,15)$ is a solution of the system.

Solution:

The only solution is $(6,10,15)$.

First we note that both $3$ and $5$ divide $8 z^{3}$, so, since $8$ is coprime to both, $3$ and $5$ must divide $z^{3}$ and hence also $z$. Let us then set $z=15 z'$, so that the first equation becomes $x y^{2}=600 z'^{3}$ with the condition $x y z'<1000 / 15<67$. If $z'$ were greater than or equal to $2$ we would have
$$x^{2} y^{2} \geq x y^{2}=600 z'^{3} \geq 600 \cdot 2^{3}=4800 .$$
This would imply $x y>69$, in contradiction with the condition $x y z'<67$. Therefore $z'=1$ and the equation becomes $x y^{2}=600$ with the condition $x y<67$. If $x$ were greater than or equal to $8$ we would have
$$x^{2} y^{2} \geq 8 x y^{2}=8 \cdot 600=4800 .$$
This would again imply $x y>69$, giving another contradiction, so we have $x<8$. Since $y^{2}=\frac{600}{x}$, we have that $\frac{600}{x}$ is the square of an integer. An easy check shows that the only possibility is $x=6$, from which we obtain $y=10$ and $z=15$. On the other hand, the triple $(6,10,15)$ is a solution of the system.