By Cauchy's inequality, we have
$$(x + y + z^2)(x + y + 1) \ge (x + y + z)^2, \qquad (6)$$
hence
$$\frac{1}{x+y+z^2} \le \frac{x+y+1}{(x+y+z)^2}.$$
Thus it suffices to show that
$$\sum_{cyc} \frac{x+y+1}{(x+y+z)^2} = \frac{2(x+y+z)+3}{(x+y+z)^2} \le 1.$$
This is equivalent to the inequality
$$(x + y + z)^2 - 2(x + y + z) - 3 \ge 0,$$
which holds for $x + y + z \ge 3$.

Equality in (6) holds if and only if $(x, y, z^2)$ and $(x, y, 1)$ are collinear, i.e., $z^2 = 1$ or, equivalently, $z = 1$. Cyclic permutation shows that equality holds if and only if $x = y = z = 1$.