Extreme values of $|z|$ when $|z^2+1|=|z-1|$

Some additional elements versus what you did.

Important to note is that $a = 2 \Re(z)$ where $\Re(z)$ stands for the real part of $z$. Indeed if $z \in S = \{z \in \mathbb C \setminus \{0\} \mid \vert z^2+1\vert=\vert z-1 \vert\} $ then (I haven't check your computations)

$$|z|^2=\frac{3\pm\sqrt{9-4(a^2+a)}}{2}.$$

And $\vert z \vert$ belongs to the interval $I= [0, \sqrt{\frac{3+\sqrt{10}}{2}}]$.

Now the question is are those values really reached?

We have $\vert z \vert = \sqrt{\frac{3+\sqrt{10}}{2}}$ for $a = 2\Re(z)= -1/2$ which represents a vertical line in the complex plane. As $\sqrt{\frac{3+\sqrt{10}}{2}} > 1/4$, we indeed have two complex satisfying the equation: the intersections of the vertical line with the circle $\vert z \vert = \sqrt{\frac{3+\sqrt{10}}{2}}$.

Let's now have a look at $\vert z \vert =0$.This would only be possible for $\sqrt{9-4(a^2+a)} = 3$, i.e. $a^2+a=0$ or $ a = 0,-1$. $a=0$ is to be ignored knowing the hypothesis $z \neq 0$. $a = -1$ cannot be if $\vert z \vert =0$. Therefore if a minimum value is attained for $\vert z \vert$, it cannot be $0$. However $0$ is a lower bound.

Another interesting question that could be studied is : is $\vert z \vert$ attaining a local minimum on $S$?