Let $a=\rho e^{i\theta_a}, b=\rho e^{i\theta_b}, c=\rho e^{i\theta_c} $. If $e^{i\eta}$ is a root of $ax^2+bx+c$, by Viète's theorem the other root is $\frac{c}{a}e^{-i\eta}$ and
$$ e^{i\eta}+\frac{c}{a}e^{-i\eta} = -\frac{b}{a}.$$
All the terms in this equation are complex numbers with unit modulus, but the sum of two complex numbers with unit modulus is still a complex number with unit modulus in very few cases, namely just the ones in which the difference of the arguments is $\frac{2\pi}{3}$. That gives that the angle between $a$ and $c$ is $\frac{2\pi}{3}$, and by a similar argument with the second polynomial we also have that the angle between $a$ and $b$ is $\frac{2\pi}{3}$, so the triangle with vertices at $a,b,c$ is equilateral and the claim follows.
From $|z_2|=|z_3|$ and $z_2=zz_3,$ we conclude $|z|=1.$ From $|z_1|=|z_3|$ and $z_1=-z_2-z_3,$ we conclude $|z+1|=1.$ If we put this together, we get $z\in\{\epsilon,\bar\epsilon\}$ where $\epsilon$ and $\bar\epsilon$ are the solutions of $z^2+z+1=0.$
We have to check if $p(z)=(-z-1)^n+z^n+1=0$ has any other solutions but $0,-1,\epsilon,\bar\epsilon.$
$0$ and $-1$ are roots of $p$ if and only if $n$ is odd. Furthermore, it can easily be verified (using the derivative of $p$) that they are simple roots in this case.
We also know that $\epsilon$ and $\bar\epsilon$ have the same multiplicity as roots of $p,$ because the coefficients of $p$ are real. If we put all this together and scale the polynomials such that the coefficients of $z^{n-1}$ match, we get the following equations that must hold if the choice of $n$ enforces the triangle to be equilateral:
\begin{eqnarray}
nz(z+1)(z^2+z+1)^{(n-3)/2} &=& (z+1)^n -z^n-1 \text{ for odd } n \\
2(z^2+z+1)^{n/2} &=& (z+1)^n +z^n + 1 \text{ for even } n
\end{eqnarray}
For odd $n\geq 5,$ the coefficient of $z^3$ in $nz(z+1)(z^2+z+1)^{(n-3)/2}$ after expansion is $\frac{(n+3)n(n-3)}{8}$, while the coefficient of $z^3$ in $(z+1)^n -z^n-1$ is $\binom n3.$ Those two match only if $n\in\{5,7\}.$
For even $n\geq 6,$ the coefficient of $z^3$ in $2(z^2+z+1)^{n/2}$ after expansion is $\frac{(n+8)n(n-2)}{24}$, while the coefficient of $z^3$ in $(z+1)^n +z^n+1$ is $\binom n3$ again. Those two do not match for any value of $n\geq 6.$
The values you have found are the only ones that enforce the triangle to be equilateral.
Best Answer
Differentiate your equation $$nz^{n-1}=\frac{d}{dz}(z-z_1)...(z-z_n)$$
All the terms formed when we differentiate vanish when we put $z=z_1$ except for $$(z-z_2)...(z-z_n)$$ which becomes $$(z_1-z_2)...(z_1-z_n)$$
So $$n=n|z_1|^{n-1}=|z_1-z_2|...|z_1-z_n|$$
Is that OK now?