Given a non-zero Lie algebra $\mathcal{L}$ over $\mathbb{C}$, we define $\mathcal{L}^2 = \big[\mathcal{L}, \mathcal{L} \big] = \big\{ [x, y]: x, y\in \mathcal{L} \big\}$, and for any $k\in\mathbb{N}$ with $k\geq 2$, define $\mathcal{L}^{k+1} = \big[ \mathcal{L}, \mathcal{L}^k \big]$. We call $\mathcal{L}$ nilpotent if there exists $n\in\mathbb{N}$ such that $\mathcal{L}^n = \{0\}$, and when $\mathcal{L}$, we call the least $n\in\mathbb{N}$ such that $\mathcal{L}^n = \{0\}$ the nil-index of $\mathcal{L}$. For instance, any abelian Lie algebra is nilpotent and their nil-index is $2$.
Now suppose $\mathcal{L}$ is a non-nilpotent and non-zero Lie algebra. Suppose $\mathcal{I}$ the family of nilpotent two-sided non-trivial ideals in $\mathcal{L}$ is non-empty. Observe that for any $I, J\in \mathcal{I}$, $I+J$ is also a non-trivial ideal and $(I+J)^3 \subseteq I^2 + J^2 \subseteq (I+J)^2 \subseteq I+J$. Hence $I+J$ is also nilpotent. If the nil-index of $I$ is $n$ and the nil-index of $J$ is $m$, then the nil-index of $I+J$ is at most $\max(n, m)+1$.
Put $\mathcal{J} = \operatorname{Span}_{\mathbb{C}} \Big( \bigcup_{I\in \mathcal{I}} I\Big)$. Clearly $\mathcal{J}$ is a two-sided ideal, although it could be the entire $\mathcal{L}$. My questions is: if $\mathcal{J}$ is a proper ideal, is it necessarily nilpotent? If not (i.e. if $\mathcal{J} = \mathcal{I}$ or if $\mathcal{J}$ is not nilpotent), can we find a maximal nilpotent ideal in $\mathcal{L}$?