Let $P$ be a finite field and $R=M_n(P[x])$ be a matrix polynomial ring. I want to prove that for every polynomial (not necessary with invertible leading term) $A(x)\in R$ such that $R\cdot A(x)$ is a maximal left ideal there exists a unique monic polynomial $B(x)\in R$ such that $$ R\cdot A(x) = R\cdot B(x) $$
Crossposted MO.
Morita equivalence for matrix algebras says $M_n(P[x])-\mathrm{Mod}\cong P[x]-\mathrm{Mod}$, so there is a bijection between the maximal left ideals (which gives irreducible modules) of $M_n(P[x])$ and those of $P[x]$.
Let's see concretely how they are corresponded. There is a $M_n(P[x])$-$P[x]$-bimodule, the standard module $P[x]^n$, which is a projective generator for both $M_n(P[x])-\mathrm{Mod}$ and $P[x]-\mathrm{Mod}$. Then, given any $M_n(P[x])$-module $M$, we associate to it the $P[x]$ module $\mathrm{Hom}_{M_n(P[x])}(P[x]^n, M)$. Conversely given any $M_n(P[x])$-module $N$, we associate to it the $M_n(P[x])$ module $P[x]^n\otimes_{P[x]} N$.
In this way, we see in particular that the regular left-module $M_n(P[x])$ gives the free $P[x]$-module $P[x]^n$ of rank $n$ and that maximal left ideals correspondes to minimal quotients of $P[x]^n$. If $I = P[x]A(x)$ is a maximal left ideal, then $M_n(P[x])/I$ corresponds under the Morita equivalence to $P[x]/(\det A(x))$. Thus $\det A(x)$ must be a irreducible polynomial in $P[x]$.
If $B(x)\in M_n(P[x])$ is a monic polynomial, then $\det B(x)$ must be of degree divisible by $n$. We see that those maximal left ideal $I$ such that $M_n(P[x])/I$ is of dimension not divisible by $n$ cannot be generated by a single monic polynomial in $M_n(P[x])$, for example, the left ideal $M_2(P[x])\begin{pmatrix}x & 0 \\ 0 & 1\end{pmatrix}$.
When a monic polynomial $B(x)$ generated a maximal left ideal $I$, we know however that $B(x)$ must be the unique one that does so. To see that, if $C(x)$ is another, then $C(x) = D(x)B(x)$ for some $D(x)\in M_n(P[x])$. We see then that $\mathrm{deg} C(x) = \mathrm{deg} B(x)$, so it is clear that $D(x) = \mathrm{I}_n$.