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.

The answer is no:

A simple C*-algebra in which $[x,y]^2$ is central for all $x,y$ is isomorphic to $M_n(\mathbf{C})$ for $n\in\{1,2\}$.

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 knows (von Neumann's bicommutant theorem) to be the strong operator topology ($\mathsf{SOT}$) closure.

Recall that an associative algebra $A$ is a PI-algebra if it satisfies some nontrivial identity $\forall x\in A^n,$ $F(x_1,\dots,x_n)=0$, where $F$ is nonzero in the free associative algebra $\mathbf{C}\langle X_1,\dots,X_n\rangle$  - $F$ can be then chosen homogeneous.

In the above setting, we have a continuous algebra homomorphism with dense image $A\to (B(H),\mathsf{SOT})$. If $A$ satisfies the identity $F$ (say of degree $d$) so is $B(H)$. This implies (by a theorem of Kaplansky, see §2.1 in Procesi's book) that $H$ has dimension $\le d/2$. Note that in the precise case in consideration, we have $F(x,y,z)=[[x,y]^2,z]$, which is clearly not an identity in $B(H)$ when $\dim(H)\ge 3$.

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 yields the complement result:

A simple C*-algebra is a PI-algebra if and only if it is isomorphic to a matrix algebra.

This follows from Kaplansky's theorem, but also directly follows if one can embed the free associative $\mathbf{C}$-algebra on countably many variables into $B(H)$ for $H$ separable infinite-dimensional.

Actually from Dixmier's book, §2.9, the intersection of kernels of irreducible representations is reduced to 0. Hence one gets a stronger result:

A C*-algebra that is a PI-algebra admits a family of finite-dimensional representations of bounded dimension, whose intersection of kernels is trivial.

(In the case of the identity $[[x,y]^2,z]$ these are of dimension $\le 2$. I'm not sure if one can strengthen the latter conclusion to a more explicit description, but in any case this is beyond what is asked, since the simple case is clear-cut now.)