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.

Note that $\mathcal{A}$ is contained in $\def\F{\mathbb{F}}\F_{q^2}$ if and only if $(q^n+1)/(q+1)$ divides $q^2-1$, and it is easy to check with a bit of case analysis that this happens only if $(n, q) = (3,2)$. In all other cases we have some $a \in \mathcal{A} \setminus \F_{q^2}$.

Let $\langle u, v\rangle = \mathrm{Tr}_{\F_{q^{2n}} / \F_{q^2}}(uv)$ denote the trace form on $\F_{q^{2n}}$, so $\langle, \rangle$ is a nondegenerate $\F_{q^2}$-bilinear form. Note that the trace of $\F_{q^{2n}} / \F_{q^2}$ restricts to the trace of $\F_{q^n} / \F_q$: indeed both are just $x + x^{q^2} + \cdots + x^{q^{2(n-1)}}$. Hence $\langle, \rangle$ restricted to $\F_{q^n}$ takes values in $\F_q$.

Let $\def\eps{\varepsilon}\eps \in \F_{q^2} \setminus \F_q$. Then $\F_{q^{2n}} = \F_{q^n}(\eps) = \F_{q^n} + \F_{q^n}\eps$.

Since $a \notin \F_{q^2}$ we can find a hyperplane containing $1$ but not $a$, so there is some $v$ such that $\langle 1, v\rangle = 0$ but $\langle a, v \rangle \neq 0$. Since $\F_{q^{2n}} = \F_{q^n} + \F_{q^n}\eps$ we have $v = x + y \eps$ for some $x, y \in \F_{q^n}$. Hence $\langle 1, x\rangle + \langle 1, y\rangle \eps = 0$ and $\langle a, x \rangle + \langle a, y \rangle \eps \neq 0$. This implies $\langle 1, x \rangle = \langle 1, y \rangle = 0$ (since both are elements of $\F_q$) but at least one of $\langle a, x\rangle$, $\langle a, y\rangle$ is nonzero.

This achieves what you want but with the roles of $=0, \neq0$ the wrong way around, i.e., $\langle 1, y\rangle = 0$ but $\langle a, y \rangle \neq 0$. Maybe there's a some way of fixing this.

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 $q^2-1$.

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.