Let $\mathbb{F}_{q^n}/\mathbb{F}_q$ be an extension of finite fields.

Is a proper quotient of $\mathbb{F}_{q^n}[x]$ considered as an $\mathbb{F}_q$-algebra always a quotient of $\mathbb{F}_q[x]$ (i.e. no extra generator is necessary)?

I think this comes down to the following. Given an irreducible $r\in \mathbb{F}_q[x]$ of degree $n$ and a positive integer $m$ can we always find $s\in \mathbb{F}_q[x]$ of degree $m$ and $t\in \mathbb{F}_q[x]$ of degree $n$ such that$$r\circ s=t^m$$ as polynomials?

For example consider $\mathbb{F}_4[x]/(x^2)$. Then$$r=x^2+x+1, \quad m=2, \quad s=x^2+1, \quad t=x^2+x+1$$and$$r\circ s=t^2=x^4+x^2+1$$so $\mathbb{F}_4[x]/(x^2)\cong\mathbb{F}_2[x]/(x^4+x^2+1)$ as $\mathbb{F}_2$-algebras.