I'm having trouble understanding what the following equality implies.
$$\frac{z_1-z_3}{z_3-z_2}=\frac{z_2-z_1}{z_1-z_3}.$$
I suspect that this means that the points form the vertices of an equilateral triangle in the complex plane, but can't prove that the points are equidistant from another, any gentle nudges?
Yes, they form an equilateral triangle.
Consider the transform $z\to t=\dfrac{z-z_3}{z_1-z_3}$. This transform is a similarity (translation, rotation, scaling) that maps $z_3$ to $0$ and $z_1$ to $1$.
In the transformed plane, the equation reduces to $$\frac{1-0}{0-t_2}=\frac{t_2-1}{1-0},$$ or $$t_2^2-t_2+1=0.$$
The solutions are obviously $$t_2=\frac{1\pm\sqrt3i}2.$$ With the segment $0-1$, they form two equilateral triangles with unit sides.
Back to the original plane, $$z_2=z_3+(z_1-z_3)\frac{1\pm\sqrt3i}{2}.$$
Double checking, $$|z_2-z_3|=|z_1-z_3|\cdot|\frac{1\pm\sqrt3i}{2}|=|z_1-z_3|,$$ $$|z_2-z_1|=|z_1-z_3|\cdot|-1+\frac{1\pm\sqrt3i}{2}|=|z_1-z_3|.$$
Let$$t=\frac{z_1-z_3}{z_3-z_2}=\frac{z_2-z_1}{z_1-z_3}.$$ We have $$t=\frac{z_1-z_3}{z_3-z_2}=\frac{z_2-z_3+z_3-z_1}{z_1-z_3}=\frac{z_2-z_3}{z_1-z_3}-1=-\frac1t-1,$$ or $$t^2+t+1=0.$$ From $$t^2+t+1=\frac{t^3-1}{t-1},$$ we deduce that $t$ is a root of $1$ and has unit module, hence $$|z_1-z_2|=|z_2-z_3|=|z_3-z_1|.$$
Here is a more geometric way to understand it. Given $\frac{a-c}{b-c} = \frac{b-a}{c-a}$, each expression tells you the directed angle between the vectors. Thus $\angle BCA = \angle CAB$ and hence $|AB| = |BC|$. But the given condition is also $(a-c)^2 = (b-a)(b-c)$ and hence $|AC|^2 = |AB| \times |BC| = |AB|^2$. Therefore $|AC| = |AB|$ and $\triangle ABC$ is equilateral.
$\arg\left(\dfrac{z_1-z_3}{z_3-z_2}\right)$ is the angle from $\overrightarrow{z_2z_3}$ to $\overrightarrow{z_3z_1}$ and $\arg\left(\dfrac{z_2-z_1}{z_1-z_3}\right)$ is the angle from $\overrightarrow{z_3z_1}$ to $\overrightarrow{z_1z_2}$.
Since $\angle z_3z_1z_2=\angle z_2z_3z_1$, $\triangle z_1z_2z_3$ is isosceles, and therefore, $\left|z_2-z_1\right|=\left|z_3-z_2\right|$.
Furthermore, since $\dfrac{\left|z_1-z_3\right|}{\left|z_3-z_2\right|}=\dfrac{\left|z_2-z_1\right|}{\left|z_1-z_3\right|}$, $\left|z_1-z_3\right|$ is the geometric mean of $\left|z_3-z_2\right|$ and $\left|z_2-z_1\right|$, which are equal. Therefore, $$ \left|z_1-z_3\right|=\left|z_3-z_2\right|=\left|z_2-z_1\right| $$ and $\triangle z_1z_2z_3$ is equilateral.
Without loss of generality, $z_2=0$ (translate by $-z_2$):
$$\frac{z_1-z_3}{z_3}=\frac{-z_1}{z_1-z_3},$$ or $$z_1^2-z_1z_2+z_3^2=0=\frac{z_1^3+z_3^3}{z_1+z_3}.$$ then $|z_1|=|z_3|$, and $|z_1-z_3|=|\sqrt{-z_1z_3}|=|z_1|=|z_3|$.
Knowing that the three points form an equilateral triangle, let us express the rotational symmetry $$z_2-z_1=(z_3-z_2)e^{\pm i2\pi/3}=(z_1-z_3)e^{\pm i4\pi/3}.$$
The given equation turns to $$e^{\mp i2\pi/3}=e^{\pm i4\pi/3},$$ an identity.
As the equation can be put in a quadratic form, we conclude that there are no more than these two solutions.
$$\frac{z_1-z_3}{z_3-z_2}=\frac{z_2-z_1}{z_1-z_3}$$
There exist $a\in\mathbb{C}, r\in\mathbb{R^+}, \theta\in[0,2\pi)$ such that $$z_3-z_2=are^{i\theta},\,\,\,\,z_1-z_3=ar^2e^{2i\theta},\,\,\,\,z_2-z_1=ar^3e^{3i\theta}$$
Also $$(z_3-z_2)+(z_1-z_3)+(z_2-z_1)=are^{i\theta}(1+re^{i\theta}+r^2e^{2i\theta})=0$$ Therefore $re^{i\theta}\not=1$ is a root of the equation $z^3=1.$ (why ?)
Therefore$$r=?,\,\,\,\,\,\theta=?$$
Hence $z_1z_2z_3$ are form a equilateral triangle.
