Let $R$ be a local Noetherian ring such that every nonzero prime ideal is maximal and every $P$-primary ideal is a power of $P$. I want to show that $R$ is a principal ideal ring.
Let $I \vartriangleleft R$, if $I=0$ it is principal. Otherwise write a primary decomposition $I = \bigcap Q_i$. Then $I = \bigcap \sqrt{Q_i}^{n_i}$ and since the radicals are nonzero and prime we have $I = \bigcap m^{n_i} = m^{\max \{n_i\}}$, therefore it suffices to show that $m$ is principal. How can I show that?
If $m=m^2$ then by Nakayama's lemma $m=0$. In this case $R$ is a field, and the result is trivial. So assume $m\ne m^2$. In this case, take $t\in m\setminus m^2$. Since you already proved that every nonzero ideal of $R$ is a power of $m$, it follows that if $(t)$ is the ideal generated by $t$ then $t=m^n$ for some $n$. Since $t\notin m^2$, the only option is $(t)=m$.