Let's work in $\mathbb{C}$.So the rational points are precisely $\mathbb{Q}[i]$. Lemma 1: if $x,y,z$ be three distinct non-collinear points in $\mathbb{Q}[i]$ then the circumcenter of $\triangle xyz$ is in $\mathbb{Q}[i]$ Click to reveal hidden textthis follow from the fact that if $t$ is the circumcenter of $\triangle xyz$ then we get two linear equations in $t,\overline t$ with coefficients in $\mathbb{Q}[i]$ Lemma 2: if $\zeta \in \mathbb{Q}[i]$ is a primitive $n^{th}$ root of unity , then $n=1,2$ or $4$ Now, suppose $\{z_1,z_2,\cdots,z_n\}$ is a regular n-gon(n$\ge$ 3) . Further suppose that $\exists$ three distinct indices $\{i,j,k\} \in \{1,2,\cdots,n\}$ such that $\{z_i,z_j,z_k\} \in \mathbb{Q}[i]$.By lemma 1. the circumcenter,i.e. $z = \frac{\sum_{i=1}^n z_i}{n}$, of the polygon must be in $\mathbb{Q}[i]$. Now consider the following map $x \mapsto \frac{x-z}{z_i-z}$.It preserves $\mathbb{Q}[i]$ as well as regularity of the polygon (because it is a composition of translation and rotation). See that $\{z_1,z_2,\cdots,z_n\}$ maps to a permutation of $\{1,\zeta,\zeta^2,\cdots,\zeta^{n-1}\}$, where $\zeta$ is primitive $n^{th}$ root of unity. See that $\{z_j,z_k\} = \{\zeta^l,\zeta^m\}, \exists \{l,m\} \in \{1,2,\cdots,n-1\}, \in \mathbb{Q}[i]$.Now by lemma 2. we must have $\{\frac{n}{gcd(n,l)},\frac{n}{gcd(n,m)}\} =1,2$ or $4$ .if $4\not | n$ then we have a contradiction and so there can't be more than two rational points or else we have $\{1,\zeta^{\frac{n}{4}},\zeta^{\frac{n}{2}},\zeta^{\frac{3n}{4}}\}\in \mathbb{Q}[i]$ and further we can't have more than four rational points.