If a left ideal $\mathscr{U} \subseteq R$ is nil, then $\mathscr{U} \subseteq rad(R)$.
proof
Let $y \in \mathscr{U}$ where $\mathscr{U}$ is nil. Then $y$ is nilpotent and $xy \in \mathscr{U}$ for any $x \in R$. It follows that $1-xy$ has an inverse $\sum_{i=0}^\infty (xy)^ii$ and since $1-xy$ has, in particular, a left inverse $y \in rad(R)$.
confusion
How is $\sum_{i=0}^\infty (xy)^ii$ even well-defined. I see the even terms will all be $0$ since $(xy)^2$ is 0 but we have no notion of convergence for such an infinite sum. How one would even verify or construct this as he inverse I am completely unclear on.
Instead of that argument you can be a bit more concrete. Let $k$ be an integer such that $(xy)^{k}=0$ (which exists because $xy$ is nilpotent). Define $$u=\sum\limits_{i=0}^{k-1}(xy)^{i}\in R,$$ and note that $$u(1-xy)=\sum\limits_{i=0}^{k-1}(xy)^{i}-\sum\limits_{i=1}^{k}(xy)^{i}=1-(xy)^{k}=1,$$ so that $u$ is a left inverse of $xy$. Since $x$ was arbitrary you can conclude that $y\in \operatorname{Rad}(R)$ and you get the desired result.