On a special type of ideal in local Artinian ring

Question

Let $(R,\mathfrak m)$ be a local Artinian ring. If $J$ is a non-zero ideal of $R$ such that $J^2=\mathfrak mJ$, then is it true that $J=\mathfrak m$ ? or at least $J^2=\mathfrak m^2$ ?

NOTE: $J^2=\mathfrak mJ\implies J^{n+1}=\mathfrak m^nJ,\forall n>1$, and since $\exists n_0>1$ such that $\mathfrak m^{n_0}=0$, so we get $J^n=0$ for some $n>1$. But I am unable to proceed further.

Answer 1

No. For instance, let $k$ be a field and let $R=k[x]/(x^3)$, with maximal ideal $\mathfrak{m}=(x)$. Then $J=(x^2)$ satisfies $J^2=\mathfrak{m}J=0$ but $J^2\neq \mathfrak{m}^2$.