It is easy to see that the triples $(0; 0; 0)$ and $(-1; -1; -1)$ satisfy the system. We will show that there are no other solutions.
From the equations
$$(x+1)(x^2+1) = y^3+1, \quad (1)$$
$$(y+1)(y^2+1) = z^3+1, \quad (2)$$
$$(z+1)(z^2+1) = x^3+1 \quad (3)$$
it follows that if $x = -1$, then, from (1) $y = -1$, and from (2) $z = -1$. Similarly, if $x = 0$, then $y = 0$ and $z = 0$. Further assume that none of $x$, $y$ and $z$ equals $-1$ or $0$.
Factor the right-hand sides of (1), (2), (3) and multiply the results:
$$(x+1)(x^2+1)(y+1)(y^2+1)(z+1)(z^2+1) = (y+1)(y^2-y+1)(z+1)(z^2-z+1)(x+1)(x^2-x+1).$$
Reducing by $(x+1)(y+1)(z+1) \ne 0$, we get
$$(x^2+1)(y^2+1)(z^2+1) = (x^2-x+1)(y^2-y+1)(z^2-z+1). \quad (*)$$
Note that if $x < -1$, then, according to (1), $y < -1$ and, according to (2), $z < -1$. In this case all factors in the right-hand side of (*) are greater than corresponding factors in the left-hand side — a contradiction.
Now factor the left-hand sides of equations (1), (2) and (3), and subtracting 1 from both sides of these equations, we obtain the equations
$$x(x^2 + x + 1) = y^3, \quad (4)$$
$$y(y^2 + y + 1) = z^3, \quad (5)$$
$$z(z^2 + z + 1) = x^3. \quad (6)$$
Note that if $x > 0$, then, according to (4), $y > 0$ and, according to (5), $z > 0$. In this case all factors in the right-hand side of (*) are less than corresponding factors in the left-hand side — a contradiction.
It remains to consider the case, when $x, y$ and $z$ belong to the interval $(-1; 0)$ and, in particular, are negative. Again, all factors in the right-hand side of (*) are greater than corresponding factors in the left-hand side. So, there are no other solutions.

Consider the values of polynomials $x^3 + x^2 + x$ and $x^3$ on the intervals $(0; +\infty)$, $(-\infty; -1)$ and $(-1; 0)$.
If $x > 0$, then $x^3 + x^2 + x > x^3 > 0$. From the equations of the system successively obtain the inequalities $y > x > 0$, $z > y > 0$ and $x > z > 0$, whence $x > z$.
If $x < -1$, then $x^3 + x^2 < 0$ and $x^2 + x > 0$, hence $-1 > x^3 + x^2 + x > x^3$. From the equations of the system we obtain the inequalities $-1 > y > x$, $-1 > z > y$ and $-1 > x > z$, and again $x > z$.
If $x \in (-1; 0)$, then $x^3 + x^2 > 0$ and $x^2 + x < 0$, therefore $-1 < x^3 + x^2 + x < x^3 < 0$. From the system we obtain the inequalities $-1 > y > x$, $-1 > z > y$ and $-1 > x > z$, whence again $x > z$.
In all three cases we got a contradiction and the two remaining variants $x = 0$ and $x = 1$ easily lead us to the triples from the answer.