Here's the differential equation:
$$\ddot{z}(t)=1-\frac{1}{(z(t)+z^\star)^2}\,u(t),$$ where $z^\star>0$ is a fixed parameter, and $u(t)$ is some arbitrary input.
I know that at equilibrium, rates of change are zero (i.e. physically no acceleration, no velocity etc.), but even if I set the above equation to zero and solve for $z,$ velocity could potentially still be non-zero, correct?
For an equilibrium, you should have a constant solution $z(t)=C$. In your equation this become: $$0=1-\frac{1}{(C+z^\star)^2}\,u(t),$$ that is $$u(t)=(C+z^\star)^2,$$ for all $t$. This is possible only if your function $u(t)$ is, itself, a constant. Maybe your confusion is when you "set the above equation equal zero". By doing so you are imposing a solution of the form $z(t)=at+b$ as $z''=0$. For this to be a solution you restrict the form of the $u(t)$ as well: $u(t)=(at+b+z^\star)^2$.