Let $(R,\mathfrak m,k)$ be an Artinian Gorenstein local ring such that $$\mu(\mathfrak m)=2, \quad\mathfrak m^2\ne 0,\quad\text{and}\quad \mathfrak m^3=0.$$ Then, is it true that every non-maximal ideal of $R$ is principal?

*Thoughts*: We have $\mathfrak m^2 \subseteq (0:_R \mathfrak m)$. Since $R$ is Artinian Gorenstein, so $\dim_k (0:_R \mathfrak m)=1$, hence $\dim_k \mathfrak m^2=1$. Thus, $\mathfrak m^2, (0:_R \mathfrak m)$ are principal ideals. For an arbitrary ideal $I\subsetneq \mathfrak m$, we have
$$\begin{array}{r@{}l}
\mu(I)
&=\dim_k(I/\mathfrak mI)\\
&\le \dim_k(\mathfrak m/\mathfrak m I)-1\\
&=\dim_k(\mathfrak m/\mathfrak m^2)+\dim_k(\mathfrak m^2/\mathfrak m I)-1\\
&\le \mu(\mathfrak m)+1-1\\
&=2,
\end{array}
$$
which is just one off from being principal.

[Also note that since $\mu(\mathfrak m) -\dim R=2$ and $R$ is Gorenstein, so a result of Serre implies $R$ is complete intersection]