Let $k$ be a field of characteristic 0 (not necessarily algebraically closed), let $G$ be a connected split reductive group over $k$ and let $\mathfrak{g}$ be the Lie algebra of $G$. 

Let $X \in \mathfrak{g}$ be a nilpotent element. Does there exist a unipotent subgroup $U$ of $G$ such that $X$ is contained in the Lie algebra of $U$ ?

If $k$ is algebraically closed this is Theorem 5.1 of <http://virtualmath1.stanford.edu/~conrad/249BW16Page/handouts/applgr.pdf>.

Here is a rough idea for a proof in the general case but I can't make the details work.

By the result over an algebraically closed field there exists a unipotent subgroup $U\_{\overline{k}}$ of $G\_{\overline{k}}$ such that $X$ is in the Lie algebra of $U\_{\overline{k}}$. Via the expoential/logarithm there is an isomorphism between $U\_{\overline{k}}$ and it's Lie algebra thus there exists $x \in U\_{\overline{k}}(\overline{k})$ whose exponential is $X$.

Since $X$ is defined over $k$ I would expect it to be also true for $x$ (as an element of $G$). Then by Theorem 3.6 (very easy in characteristic 0) of the Conrad's notes, $x$ will be an element of a unipotent subgroup $U$ of $G$ and we would deduce that $X$ is contained in the Lie algebra of $U$.

Since I work in characteristic 0, I imagine there might be a much simpler way. Also I don't think the split reductive hypothesis is necessary (at least I don't use it in my "proof idea").