Yes, these are the only solutions. Let $\alpha = e^{p \pi i/n}$ and $\beta = e^{q \pi i/n}$. So the equation is
$$4 \left( \frac{\alpha+\alpha^{-1}}{2} \right) \left( \frac{\beta+\beta^{-1}}{2} \right) + 2 \left( \frac{\alpha+\alpha^{-1}}{2} \right) + 2 \left( \frac{\beta+\beta^{-1}}{2} \right) =1.$$

We rewrite this as
$$\alpha \beta + \alpha + \alpha \beta^{-1} + \beta + \beta^{-1} + \alpha^{-1} \beta + \alpha^{-1} + \alpha^{-1} \beta^{-1} = 1.$$
$$\alpha \beta + \alpha + \alpha \beta^{-1} + \beta + 1+\beta^{-1} + \alpha^{-1} \beta + \alpha^{-1} + \alpha^{-1} \beta^{-1} = 2.$$
$$(\alpha+1+\alpha^{-1}) (\beta + 1 + \beta^{-1}) = 2. \quad \label{1}\tag{$\ast$}$$
So we want to solve equation \eqref{1} in roots of unity. The conjecture in the question, which we will prove, is that the only solutions are $(\alpha, \beta) = (e^{\pm i \pi/3}, e^{\pm i \pi/4})$ or $(e^{\pm i \pi/4}, e^{\pm i \pi/3})$, so $(\alpha+1+\alpha^{-1}, \beta + 1 + \beta^{-1}) = (2,1)$ or $(1,2)$.

This solution uses the language of algebraic number theory, which I'm afraid may not be familiar to the original poster, but I can't figure out how to do this in a more elementary way.

Let $R = \mathbb{Z}[e^{\pi i/n}]$. Let $\mathfrak{p}$ be a prime of $R$ lying over the prime $2$ in $\mathbb{Z}$, and let $v : R \to \mathbb{Q}$ be the $\mathfrak{p}$-adic valuation, normalized so that $v(2)=1$. So equation \eqref{1} gives
$$v(\alpha+1+\alpha^{-1}) + v(\beta+1+\beta^{-1}) = 1. \label{2}\tag{$\clubsuit$}$$

**Lemma** Let $\eta$ be a primitive $m$-th root of unity. Then
$$v(\eta+1+\eta^{-1}) = \begin{cases} \infty & m = 3 \\ 1/2^k & m = 3 \cdot 2^{k+1} \ \text{for} \ k \geq 0 \\ 0 & \text{otherwise} \end{cases}.$$

**Proof:** The case $\eta=1$ (so $m=1$) is easy to check by hand, so we assume that $\eta \neq 1$ from now on.

We have
$$\eta+1+\eta^{-1} = \eta^{-1} \frac{\eta^3-1}{\eta-1}$$
so
$$v(\eta+1+\eta^{-1}) = v(\eta^3-1) - v(\eta-1). \label{3}\tag{$\diamondsuit$}$$
If $\omega$ is a primitive $\ell$-th root of unity, then
$$v(\omega-1) = \begin{cases} \infty & \ell=1 \\ 1/2^k & \ell = 2^{k+1} \\ 0 & \text{otherwise} \end{cases}. \label{4} \tag{$\heartsuit$}$$
(See, for example, Chapter 8 in Washington's *Introduction to Cyclotomic Fields.*)

Since $\eta$ is a primitive $m$-th root of unity, $\eta^3$ is a primitive $m/\text{GCD}(m,3)$-th root of unity, so combining \eqref{3} and \eqref{4} gives the claim. $\square$

Now, plug the Lemma into \eqref{2}. The only ways to write $1$ as a sum of numbers in $\{ \infty, 1, 1/2, 1/4, \cdots, 0 \}$ are $1+0$, $0+1$ and $1/2+1/2$.

If we are in the $1+0$ case, then $\alpha$ must be a primitive $6$-th root of unity, so $\alpha+1+\alpha^{-1} = 2$ and $\beta + 1 + \beta^{-1} = 1$, giving that $\beta$ is $\pm i$. Similarly, in the $0+1$ case, $\alpha$ is $\pm i$ and $\beta$ is a primitive $6$-th root of unity.

Finally, we are left with the $1/2+1/2$ case. In this case, $\alpha$ and $\beta$ should be primitive $12$-th roots of unity. This means that $\alpha+1+\alpha^{-1}$, and $\beta+1+\beta^{-1}$ should be $1 \pm \sqrt{3}$. We get a near miss: $(1+\sqrt{3})(1-\sqrt{3})$ is $-2$, not $2$, so this solution to \eqref{2} doesn't yield a solution to \eqref{1}.

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