Question: Can anyone prove that a [Weyl Algebra](http://en.wikipedia.org/wiki/Weyl_algebra) is not isomorphic to a matrix ring over a division ring?

Answer: 
A different proof would be to show that a Weyl algebra is not semisimple, that is, that it is not a direct sum of simple submodules as a left module over itself. However, note that there is an infinite descending chain of left submodules of a Weyl algebra given by 
$A\_n\supseteq A\_nd\supseteq A\_nd^2\supseteq A\_nd^3\supseteq...$
where $d$ is any non-invertible element. A direct sum of a finite number of simple modules can't have an infinite descending chain of submodules. Then, by the converse of Artin-Wedderburn, $A\_n$ is not a direct sum of matrix algebras over a divsion ring.

Of course, showing this sequence of submodules never stabilizes can be done by looking at the associated graded algebra, and noting that the $\overline{A\_nd^n}$ are always distinct there. However, then this answer starts getting closer to David's answer, so maybe this wasn't a truly different proof.
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.