The answer is no.
Let $A$ be a C*-algebra. Let $H$ be an irreducible representation of $A$. By irreducibility its commutant is reduced to scalars, and hence its double commutant is equal to the whole algebra $B(H)$, and the double commutant is know to be the strong operator topology closure. In particular, if $A$ is a PI-algebra (i.e., satisfies some nontrivial identity $F(x_1,\dots,x_n)=0$, where $F$ is nonzero in the free associative algebra $\mathbf{C}\langle X_1,\dots,X_n\rangle$- $F$ can be then chosen homogeneous of degree $d$), so is $B(H)$, which implies (by a theorem of Kaplansky, see §2.1 in Procesi's book) that $H$ has dimension $\le d/2$.
Now if $A$ is a simple C*-algebra, not reduced to scalars, then $A$ admits a nontrivial irreducible representation (cf §2.5 in Dixmier's book), in a Hilbert space $H$. Since $A$ is simple, this is a faithful representation. If $A$ is a PI-algebra, this forces, by the above, $H$ to be finite dimensional, so $A$ is a matrix algebra.
This applies in your particular setting: if $A$ is such that $[x,y]^2$ is scalar for all $x,y$, then it satisfies the homogeneous identity $[[x,y]^2,z]$. So $A\simeq M_n(\mathbf{C})$ for some $n$. Since $M_3(\mathbf{C})$ does not satisfy this identity (choose $x,y$ with $[x,y]=\mathrm{diag}(2,-2,0)$, whose square is not central), we deduce $n\le 2$. Note that in this particular case we don't use Kaplansky's theorem, but only the bicommutant theorem.