Rewrite the equilateral condition $z_1^2+z_2^2+z_3^2=z_1z_2+z_2z_3+z_3z_1$ as
$$z_3^2 -(z_1+z_2)z_3 + z_2^2+z_3^2-z_1z_2=0$$
which is quadratic in $z_3$, yielding
$$z_3= e^{i\frac\pi3}z_1 + e^{-i\frac\pi3}z_2$$
Substitute $z_3$ into the line equation below for the hypotenuse
$$\frac{z+\bar z}{10}+ \frac{z-\bar z}{4\sqrt3 i}=1$$
obtained from the given side lengths, leading to
$$\frac7{20}|z_1 |+ \frac{11\sqrt3}{60}|z_2 |=1
$$
where $z_1=\bar z_1$ and $z_2= -\bar z_2$ are used. Then, the area of the equilateral triangle is
$$A= \frac{\sqrt3}{4} |z_2-z_1|^2
= \frac{\sqrt3}{4}( |z_2|^2 +|z_1|^2 )\ge \frac{\sqrt3}{4} \frac{\left(\frac7{20}|z_1 |+ \frac{11\sqrt3}{60}|z_2 |\right)^2 }{\left(\frac7{20}\right)^2
+ \left(\frac{11\sqrt3}{60}\right)^2 }= \frac{75\sqrt3}{67}
$$
where the C-S inequality is applied.
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
Hint; The triangle in the complex plane whose verticies are the origin and the points $z_{1}$ and $z_{2}$ is equalateral if and only if \begin{equation} z_{1}+z_{2} = z_{1}z_{2} \end{equation} (Further hint: The reasoning behind this is that the distance from the origin to $z_{1}$ is $\sqrt{z_{1}\bar{z_{1}}}$ etc...)