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,
$$|z_2-z_3|=|z_1-z_3|\cdot|\frac{1\pm\sqrt3i}{2}|=|z_1-z_3|,$$
$$|z_2-z_1|=|z_1-z_3|\cdot|-1+\frac{1\pm\sqrt3i}{2}|=|z_1-z_3|.$$