Assuming that ring morphisms have to take identity elements to identity elements, I think that one can show even more for the first Weyl algebra (the proof for higher order Weyl algebras is analogous), with very basic methods;

Claim:
For any positive integer $n$ and any unital ring $R$ it is impossible to have a ring morphism

$$ \phi : \mathbb{C}\langle x,y\rangle/(yx-xy-1) \to M_n(R).$$

Proof:
Seeking for a contradiction, suppose that there is some $n$ and some unital ring $R$ such that $\phi$ exists. Denote the images in $M_n(R)$ of $x$ respectively $y$, under $\phi$, by $A:=\phi(x)$ respectively $B:=\phi(y)$. The image of $1$ will be the identity matrix $I$.

Consider the element $yx-xy=1$, the image of which, under $\phi$, is equal to

$$ BA-AB=\phi(yx-xy)=\phi(1)=I. $$

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 $tr(I)=n$. This is a contradiction.

Corollary of the above proof:
The same claim holds if we replace $M_n(R)$ 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 $M_n(R)$ with $R=\mathbb{C}$, because $M_n(\mathbb{C})$ 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.

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 take identity elements to identity elements, we can show the following for the $n$:th Weyl algebra, 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:
There does not exist a positive integer $m$ and an associative, commutative and unital ring $R$ such that there is a ring morphism

$$ \phi : A_n \to M_m(R).$$

Proof:
Seeking for a contradiction, suppose that there is some $m$ and some associative, commutative and unital ring $R$ such that $\phi$ exists. Denote the images in $M_m(R)$ of $x_1$ respectively $y_1$, under $\phi$, by $A:=\phi(x_1)$ respectively $B:=\phi(y_1)$. The image of $1$ will be the identity matrix $I$.

Consider the element $y_1x_1-x_1y_1=1$, the image of which, under $\phi$, is equal to

$$ BA-AB=\phi(y_1x_1-x_1y_1)=\phi(1)=I. $$

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 $tr(I)=m$. This is a contradiction.

Corollary of the above proof:
The same claim holds if we replace $M_m(R)$ 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 $M_m(R)$ with $R=\mathbb{C}$, because $M_m(\mathbb{C})$ 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.

Assuming that ring morphisms have to take identity elements to identity elements, I think that one can show even more for the first Weyl algebra, with very basic methods;

Claim:
For any positive integer $n$ and any unital ring $R$ it is impossible to have a ring morphism

$$ \phi : \mathbb{C}\langle x,y\rangle/(yx-xy-1) \to M_n(R).$$

Proof:
Seeking for a contradiction, suppose that there is some $n$ and some unital ring $R$ such that $\phi$ exists. Denote the images in $M_n(R)$ of $x$ respectively $y$, under $\phi$, by $A:=\phi(x)$ respectively $B:=\phi(y)$. The image of $1$ will be the identity matrix $I$.

Consider the element $yx-xy=1$, the image of which, under $\phi$, is equal to

$$ BA-AB=\phi(yx-xy)=\phi(1)=I. $$

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 $tr(I)=n$. This is a contradiction.

Corollary of the above proof:
The same claim holds if we replace $M_n(R)$ 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 $M_n(R)$ with $R=\mathbb{C}$, because $M_n(\mathbb{C})$ 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.

Assuming that ring morphisms have to take identity elements to identity elements, I think that one can show even more for the first Weyl algebra (the proof for higher order Weyl algebras is analogous), with very basic methods;

Claim:
For any positive integer $n$ and any unital ring $R$ it is impossible to have a ring morphism

$$ \phi : \mathbb{C}\langle x,y\rangle/(yx-xy-1) \to M_n(R).$$

Proof:
Seeking for a contradiction, suppose that there is some $n$ and some unital ring $R$ such that $\phi$ exists. Denote the images in $M_n(R)$ of $x$ respectively $y$, under $\phi$, by $A:=\phi(x)$ respectively $B:=\phi(y)$. The image of $1$ will be the identity matrix $I$.

Consider the element $yx-xy=1$, the image of which, under $\phi$, is equal to

$$ BA-AB=\phi(yx-xy)=\phi(1)=I. $$

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 $tr(I)=n$. This is a contradiction.

Corollary of the above proof:
The same claim holds if we replace $M_n(R)$ 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 $M_n(R)$ with $R=\mathbb{C}$, because $M_n(\mathbb{C})$ 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.