If $\sum\limits_{n=1}^{\infty}e_n x^n = 0$ always implies $\sum\limits_{n=1}^{\infty}e_n a_n = 0$, then $(a_n) = (x^n)$?

Question

Let $x \in \left[\frac{2}{\sqrt{5}+1}, 1\right)$ and assume the real sequence $(a_n)$ satisfies

1. $a_1 = x,$

2. For every sequence $(e_n) \subset \left\{-1, 0, 1 \right\}$ with $\sum\limits_{n=1}^{\infty}e_n x^n = 0$ we have $\sum\limits_{n=1}^{\infty}e_n a_n = 0.$

How can I show that this implies $(a_n) = (x^n)$?

I can see why it's necessary to restrict the interval from which $x$ can be taken (For large/small enough $x$ the only way to get $\sum\limits_{n=1}^{\infty}e_n x^n = 0$ is to have $e_n = 0$ for all $n$), but I have no idea how to show the implication to be true in the given interval.

As $\frac1x$ has to be between $1$ and $\frac{\sqrt{5}+1}{2}$, there must be some connection to the golden ratio, but I don't see it yet.

Any help is appreciated!

EDIT: A proof was posted on AoPS a couple of months after I asked this question here.