matrices - Radical of $mathfrak{gl}_n$

I find it intuitive enough that the radical of $\mathfrak{gl}_n\mathbb F$ is the scalar matrices, but I have trouble finding an easy, but complete proof:

Proof. Let $\mathfrak s$ denote the scalar matrices. Clearly $\mathfrak s\subset\mathrm{rad}(\mathfrak{gl}_n\mathbb F)$. Suppose that $\mathrm{rad}(\mathfrak{gl}_n\mathbb F)$ is generated by more than one element, so that $X\in\mathrm{rad}(\mathfrak{gl}_n\mathbb F)$, but $X\notin\mathfrak s$. We can change basis such that $\mathrm{rad}(\mathfrak{gl}_n\mathbb F)$ is upper-triangular. (The change of basis leaves the scalar matrices invariant.) Then $X$ is upper-triangular.

I want to conclude that there exists a $Y\in\mathfrak{gl}_n\mathbb F$ s.t. $[X,Y]$ is not upper-triangular. How can I see that $Y$ always exists, except by waving hands?