Let $R$ be a commutative finite local ring of order $p^n$ ($p$ is a prime and $1\in R$). I'm struggling with the following two basic questions:
(a) Is it true that $x^n=0$ for every non-unit $x\in R$ ?
(b) Is there exists a nilpotent element $x\in R$ such that $x^{n-1}\neq0$ ?
My guess is that the answer to both questions is false, but unfortunately I can not find any appropriate counterexamples.
I will try to answer your first question.
In any commutative local ring $R$ of order $p^n$ it is true that $x^n$=0 for every non-unit $x\in R$.
Let $R^* $ be the set of units of $R$ and $M$ be the maximal ideal of $R$. Then $|M|=p^k$ for some non-negative integer $k<n$ and $|R^*|=p^n-p^k=p^k(p^{n-k}-1)\geq p-1$.
Let $x\in M$. Let us prove that $x^n=0$.
Note first that if $x,\ldots,x^s$ are all nonzero, then $$ a_1x+\ldots+a_sx^s=0,\ a_i\in R^*\cup\{0\}\ \Rightarrow\ a_1=\ldots=a_s=0. $$ Indeed, if $a_1x+\ldots+a_sx^s=0$ and $a_1\neq0$, then $x(a_1+\ldots+a_sx^{s-1})=0\ \Rightarrow\ x=0$.
It follows that elements of the form $a_1x+\ldots+a_sx^s=0,\ a_i\in R^*\cup\{0\}$ are at least $p^s$.
So $s<n$, and then $x^n=0$.