Nilpotency is not a radical property, because a ring does not, in general, contain a largest nilpotent ideal.

Let $k$ be a field, and let $S = k[X_1, X_2, \dotsc]$ be the ring of polynomials over $k$ in infinitely many variables $X_1, X_2, \dots$ Let $I$ be the ideal of $S$ generated by $\{X_n^{n+1} \mid n \in \mathbb(N)\}$ Let $R = S/I$ Note that $R$ is commutative.

For each $n$ let $A_n = \sum_{k=1}^n RX_n$ Let $A = \bigcup A_n = \sum_{k = 1}^\infty RX_n$

Then each $A_n$ is nilpotent, since it is the sum of finitely many nilpotent ideals (see proof here). But $A$ is nil, but not nilpotent. Indeed, for any $n$ there is an element $x \in A$ such that $x^n \neq 0$ namely $x = X_n$ and so we cannot have $A^n = 0$

So $R$ cannot have a largest nilpotent ideal, for this ideal would have to contain all the ideals $A_n$ and therefore $A$