More generally, let $\def\F{\mathbb{F}}\chi : \F_{q^n} \to \F_q$ be a nontrivial character (such as the trace map). Suppose $a \in \F_{q^n}$ has the property that $\chi(x) = 0$ implies $\chi(ax)=0$ for all $x \in \F_{q^n}$. Every character of $\F_{q^n}$ has the form $\chi(vx)$ for some $v \in \F_{q^n}$, so it follows that all characters have the same property. This implies $a \in \F_q$, since otherwise there is a hyperplane containing $1$ but not $a$.
So, in your situation, the question is whether $(q^n+1)/(q+1)$ divides $q^2-1$, and it is easy to check with a bit of case analysis that $(n, q) = (3,2)$ is the only solution.