Let $R$ be an arbitrary ring with maximal ideal $I$, such that the quotient ring $\frac{R}{I}$ is finite. For each integer $i$, we define the quotient ring number as below: $$N_i(R,I)=\sum_{a\in \frac{R}{I}}{a^i}.$$
It is easy to see that if $\frac{R}{I}=\mathbb{F}_q$ be a finite field with $q$ elements, then we have two case for $N_i(R,I)$. Actually, if $i$ is not a multiple of $q-1$, this summation is zero and if $i$ is a multiple of $q-1$, this summation is $-1$. So, for commutative ring with maximal ideal $I$, we have: $$N_i(R,I)\in \{0,-1\}.$$
Now, my questions is:
Is there a non-commutative ring $R$ with maximal ideal $I$ such that we have $N_i(R,I)\in \{0,-1\}$?
Also, for arbitrary non-commutative ring $R$ with maximal ideal $I$, what can be the maximum number of elements of the set $\{N_i(R,I); i\in \mathbb{N}\}$?