This is not an answer to the original question. However, it is related and I think that it is worth mentioning.
Assuming that ring morphisms have to take identity elements to identity elements, I think that one we can show even more the following for the first $n$:th Weyl algebra(the proof for higher order Weyl algebras is analogous), with very basic methods;.
Choose an arbitrary positive integer $n$ and put $A_n := \mathbb{C}\langle x_1,\ldots,x_n,y_1,\ldots,y_n \rangle / I$ where $I$ is the ideal generated by the elements $y_1x_1-x_1y_1-1,\ldots,y_nx_n-x_ny_n-1$ and $x_i x_j-x_jx_i, y_i y_j - y_j y_i$ for $i,j \in 1,\ldots,n$.
Claim:
For any
There does not exist a positive integer $n$ m$ and any an associative, commutative and unital ring $R$ it such that there is impossible to have a ring morphism
$$ \phi : \mathbb{C}\langle x,y\rangle/(yx-xy-1) A_n \to M_n(R).$$M_m(R).$$
Proof:
Seeking for a contradiction, suppose that there is some $n$ m$ and some associative, commutative and unital ring $R$ such that $\phi$ exists. Denote the images in $M_n(R)$ M_m(R)$ of $x$ x_1$ respectively $y$, y_1$, under $\phi$, by $A:=\phi(x)$ A:=\phi(x_1)$ respectively $B:=\phi(y)$. B:=\phi(y_1)$. The image of $1$ will be the identity matrix $I$.
Consider the element $yx-xy=1$, y_1x_1-x_1y_1=1$, the image of which, under $\phi$, is equal to
$$ BA-AB=\phi(yx-xy)=\phi(1)=IBA-AB=\phi(y_1x_1-x_1y_1)=\phi(1)=I. $$
Hence the matrices $A$ and $B$ have to satisfy $BA-AB=I$. Taking the trace of the left hand side of this equality yields
$$ tr(BA-AB)=tr(BA)-tr(AB)=tr(AB)-tr(AB)=0 $$
whereas the trace of the right hand side is equal to $tr(I)=n$. tr(I)=m$. This is a contradiction.
Corollary of the above proof:
The same claim holds if we replace $M_n(R)$ M_m(R)$ by any unital Banach algebra.
This is easily seen by using the following well-known fact:
The identity element of a unital Banach algebra can not be a commutator,
i.e. $ab-ba\neq 1$ for any elements $a,b$ of the Banach algebra.
This applies to the case $M_n(R)$ M_m(R)$ with $R=\mathbb{C}$, because $M_n(\mathbb{C})$ M_m(\mathbb{C})$ is a unital C*-algebra and in particular a unital Banach algebra. So in this particular case, this gives us an alternative way of making the desired conclusion without using the trace.
