Differing definitions of Cartan subalgebras

This equivalence is proposition 4 in book VII, §2 of Bourbaki's treatise on Lie Groups and Lie Algebras. The direction you're interested in is proved via a straightforward application of Engel's Theorem:

(Proof for "$\mathfrak h$ nilpotent and self-normalising" $\Rightarrow$ "$\mathfrak{h}=\mathfrak g_0$".) There is a natural Lie algebra action of $\mathfrak{h}$ on the quotient space $V:=\mathfrak{g}_0/\mathfrak{h}$ (via adjoint). If $V \neq 0$, since $\mathfrak h$ is nilpotent, Engel's theorem guarantees that there is a $0 \neq v \in V$ such that $\mathfrak{h} \cdot v =0$. Unravelling the definitions, this means any representative $x \in \mathfrak{g}_0 \setminus \mathfrak h$ of $v$ would be in the normaliser of $\mathfrak h$, but we had assumed $\mathfrak h$ to be self-normalising, contradiction. So $V = 0$ which proves the claim, which obviously is stronger than what you seek.

As pointed out in the other answer, if $\mathfrak{g}$ is semisimple, one can even prove that $\mathfrak h = \mathfrak g_0$ is abelian; the argument above however works in any (finite-dimensional) Lie algebra $\mathfrak g$, and obviously here in general $\mathfrak h = \mathfrak g_0$ need not be abelian.

For completeness, the other direction (which in Bourbaki is hidden in a rabbit hole of references, in loc. cit. §1 proposition 10) goes like this:

(Proof for "$\mathfrak h$ nilpotent and $\mathfrak h = \mathfrak g_0$" $\Rightarrow$ "$\mathfrak{h}$ is self-normalising".) Let $x$ be in the normaliser of $\mathfrak g_0$. For any $y \in \mathfrak h$, we have $z:=ad_y(x) \in \mathfrak h$ (by definition of normaliser and $\mathfrak h = \mathfrak g_0$). But then because $\mathfrak h$ is nilpotent (or by definition of $\mathfrak g_0$), there is $n \in \mathbb N$ (depending on $y$) such that $0=(ad_y)^n(z) = (ad_y)^{n+1}(x)$. But the existence of such $n$ for each $y \in \mathfrak h$ implies by definition that $x \in \mathfrak{g}_0$. We have thus shown that $\mathfrak{g}_0$ is self-normalising.