Given, $\quad\left|z^2-1\right|=|z|^2+1$
Let
$$
z=x+i y
$$
$$
\begin{aligned}
& \Rightarrow \quad\left|(x+i y)^2-1\right|=|x+i y|^2+1 \\
& \Rightarrow\left|x^2-y^2+2 i x y-1\right|=\left(x^2+y^2\right)+1 \\
& \Rightarrow\left|\left(x^2-y^2-1\right)+2 i x y\right|=\left(x^2+y^2+1\right) \\
& \Rightarrow \sqrt{\left(x^2-y^2-1\right)^2+4 x^2 y^2}=x^2+y^2+1 \\
& \Rightarrow\left(x^2-y^2\right)^2+1-2\left(x^2-y^2\right)+4 x^2 y^2 \\
& =\left(x^2+y^2+1\right)^2 \\
& =x^4+y^4+2 x^2 y^2+1+2 x^2+2 y^2 \\
& \Rightarrow-2 x^2 y^2-2 x^2+4 x^2 y^2=2 x^2 y^2+2 x^2 \\
& \Rightarrow \quad-2 x^2=2 x^2 \\
& \Rightarrow \quad 4 x^2=0 \Rightarrow x=0 \\
& \therefore \quad z=x+i y=0+i y \\
& \Rightarrow \quad z=i y \quad \Rightarrow \quad(x, y)=(0, y) \\
&
\end{aligned}
$$
Hence, $z$ lies on the imaginary axis.