I an writing this as an answer. It is indeed the case that the graded ring of a local Artinian complete intersection ring can fail to be a local Artinian complete intersection ring. Here is a direct argument instead.

By hypothesis, $R$ is a quotient $A/I$ of a complete local ring $(A,\mathfrak{n},k)$ that is regular, where $I$ is generated by an $A$-regular sequence in $\mathfrak{n}^2$. In particular, the induced $k$-linear map from $\mathfrak{n}/\mathfrak{n}^2$ to $\mathfrak{m}/\mathfrak{m}^2$ is an isomorphism. Let $(x_1,\dots,x_d)$ be an ordered $d$-tuple of elements of $\mathfrak{n}$ that map to a $k$-basis for $\mathfrak{n}/\mathfrak{n}^2$.

Thus, the ideal $I$ is generated by a regular sequence of length $d$ inside $\mathfrak{n}^2$. Now consider the $d(d+1)/2$ $k$-linearly independent elements $x_ix_j\in \mathfrak{n}^2/\mathfrak{n}^3$. The $R$-module $\mathfrak{m}^2/\mathfrak{m}^3$ equals the quotient of the $A$-module $\mathfrak{n}^2/\mathfrak{n}^3$ by $I/(I\cap \mathfrak{n}^3)$. Also $I/(I\cap \mathfrak{n}^3)$ is a quotient of $I/I\cdot \mathfrak{n}$, which is a $k$-vector space of dimension $d$. Thus, the $k$-vector space dimension of $\mathfrak{m}^2/\mathfrak{m}^3$ is at least $d(d+1)/2 - d = d(d-1)/2$.

Thus, if the socle of $A$ is $\mathfrak{m}^2$, so that $\mathfrak{m}^2/\mathfrak{m}^3$ has $k$-vector space dimension equal to $1$, then either $d=1$ and $I\subset \mathfrak{n}^3$ or $d=2$ and $I\cap \mathfrak{n}^3$ equals $I\cdot \mathfrak{n}$, i.e., the $2$ minimal generators of $I$ are elements of $\mathfrak{n}^2$ whose images in $\mathfrak{n}^2/\mathfrak{n}^3$ are $k$-linearly independent.