All of them can be described as follows. Let $(A,m)$ be any local ring (commutative). By going modulo $m^3$ we may assume that $m^3=0$. Now $m^2$ is a vector space over $k=A/m$ and let $I\subset m^2$ be any codimension one $k$-subspace. Then $I$ is an ideal in $A$ and $A/I$ will have the property you need and any such looks like this.

All of them can be described as follows. Let $(A,m)$ be any local ring (commutative). By going modulo $m^3$ we may assume that $m^3=0$. Now $m^2$ is a vector space over $k=A/m$ and let $I\subset m^2$ be any codimension one $k$-subspace. Then $I$ is an ideal in $A$ and $A/I$ will have the property you need and any such looks like this.

As Pham pointed out, my answer was not correct. So, here is another attempt. Clearly, we may assume that $R$ is graded, since $m^3=0$. Thus, $R=k\oplus V\oplus k$ where $V\cong m/m^2$ and the last $k=m^2$. To make sure that $m^2$ is the unique maximal ideal, the only condition we need to assume seems to be the natural multiplication map $S^2V\to k$ is non-degenerate. So, all such rings seem to come from starting with a vector space $V$ and a non-degenerate symmetric bilinear form on $V$, which gives a commutative ring structure on $R$ as above.