It is a basic fact that a set of all nilpotent elements in a commutative ring is an ideal. Suppose that $A$ is a noncommutative ring, $I$ is a two-sided ideal generated by all nilpotent elements and $I_r$ is a right ideal generated by all nilpotent elements. I wonder whether one of the following cases is possible for some ring $A$:
- $I$ is a proper non-zero ideal in $A$ (answered in a post below and in a comment )
- $I_r$ is a proper subset of $I$
- Both of the statements above hold.
I think I have a partial answer for 1. Consider the ring $$ R = \{ a + bi + cj + dk : a,b,c,d \in \mathbb{Z} / m \mathbb{Z}\},$$ where $m$ is a composite integer such that the power of some of its prime factors in its factorization in primes is greater than $1$, and $i,j,k$ follow these rules: $$i^2 = j^2 = k^2 = -1, ij = k, jk = i, ki = j.$$ It follows that $R$ is a non-commutative ring. If $p_1, \dots, p_k$ are all the distinct prime factors of $m$, then $p_1 p_2 \cdots p_k$ is a non-zero element in $R$ that is nilpotent (just raise it to the greatest of the powers of these primes in the factorization of $m$), thus, if some element of $R$ is is not in the two-sided ideal generated by the nilpotent elements of $R$ (possibly the identity $1$), your first assertion would hold.