= Solution
If $\mathfrak g$ is nilpotent, every <Lie subalgebra> $\mathfrak a$ is nilpotent, since $\mathfrak a^j\subseteq\mathfrak g^j$ in the respective <lower central series of a Lie algebra>. A two-dimensional nilpotent algebra has nonzero centre by part (i). If $0\ne z$ is central, complete it to a <basis> $(z,u)$; every bracket of these <basis> elements is zero. Thus $\mathfrak a$ is abelian.
Conversely, suppose $\mathfrak g$ is not nilpotent. By <Engel's theorem>, some $\operatorname{ad}(x)$ is not nilpotent. Since the <field> is algebraically closed, it has an <eigenvalue> $\lambda\ne0$ with <eigenvector> $y\ne0$. Thus
$$
[x,y]=\lambda y.
$$
The vectors $x,y$ are independent: otherwise their bracket would be zero. Their span is therefore a two-dimensional <Lie subalgebra> and it is nonabelian. This contradicts the hypothesis. Hence
$$
\boxed{\mathfrak g\text{ is nilpotent}\iff
\text{every two-dimensional subalgebra is abelian}.}
$$
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