Let $p$ be a prime integer and $R$ be a finite local ring. Assume that $p||R^\times|$. Then by Cauchy's Theorem, there always exists a primitive $p$ root of unity in $R^\times$. Here $R^\times$ is a unit group of $R$.
Is it possible to find a finite local ring $R$ which is not a finite field and a prime integer $p>2$ dividing $|R^\times|$ such that there exists a primitive $p$ root of unity $u \in R^\times $ for which $u - 1$ is still a unit in $R$?
Sure: just take a finite field and then add some nilpotents to get a local ring that is not a field. For instance, let $F$ be a finite field such that $p$ divides $|F^\times|$ and then let $R=F[x]/(x^2)$.
You can also get examples by taking $\mathbb{Z}/q^n$ where $q$ is a prime such that $p\mid q-1$. For instance, if $R=\mathbb{Z}/49$, then there are primitive cube roots of unity $u=18$ and $u=30$ such that $u-1$ is also a unit.