Every solvable Lie algebra is nilpotent

Alternation and bilinearity determine all brackets from $[x,y]=y$, and Jacobi holds because it is enough to check basis triples, where either two entries coincide or the inner bracket is a scalar multiple of $y$. Thus $\mathfrak{a}$ is a Lie algebra. Its derived algebra is ${\mathfrak{a}}^{(1)}=ky$, and ${\mathfrak{a}}^{(2)}=[ky,ky]=0$, so it is solvable by [L1].

Its lower central series satisfies ${\gamma }_2(\mathfrak{a})=[\mathfrak{a},\mathfrak{a}]=ky$ and, whenever $r\ge 2$ and ${\gamma }_r=ky$, ${\gamma }_{r+1}=[\mathfrak{a},ky]=ky$ because $[x,y]=y\mathrm{\ne }0$. Hence every term from ${\gamma }_2$ onward is $ky$, so the series never reaches zero and $\mathfrak{a}$ is not nilpotent by [L2]. This explicit solvable nonnilpotent witness refutes the statement over every field and uses no choice.