Nilpotent elements of Lie algebra and unipotent groups 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").
 A: Indeed there's a much simpler way in characteristic zero, with $G$ an arbitrary $k$-defined linear algebraic group.
Let $X$ be nilpotent. Fix a faithful $k$-defined linear representation $\rho$ of $G$ and let $\rho'$ be the corresponding representation of $\mathfrak{g}$.
Now $\rho'(X)$ being nilpotent, it preserves a complete flag (defined over $k$) in the linear space $\mathfrak{g}$. Let $H$ be the subgroup of $G$ preserving this flag (for the representation $\rho$), and acting as identity on all successive 1-dimensional quotients. Then, since $\rho$ is faithful, $H$ is a $k$-defined unipotent subgroup of $G$, whose Lie algebra $\mathfrak{h}$ is the subalgebra of $\mathfrak{g}$ of those element preserving the flag (for $\rho'$), and acting as zero on all successive 1-dimensional quotients. Hence $X\in\mathfrak{h}$.
A: This is too complicated, but it's the first approach that comes to mind.  I take Conrad's ‘dynamical’ approach to parabolic subgroups, thinking instead of cocharacters defining them.
As you have observed, a nilpotent element lies in the Lie algebra of the unipotent radical of some parabolic subgroup over the algebraic closure, hence is strictly contracted by some cocharacter defined over the algebraic closure; so the closure of its orbit contains $0$.  It follows from Corollary 4.3 of Kempf - Instability in invariant theory (MSN) that it is actually strictly contracted by some rational cocharacter, hence lies in the Lie algebra of the unipotent radical of the corresponding rational parabolic subgroup.  (See also §2.5 of Adler and DeBacker - Some applications … (MSN).)
EDIT:  The analogous result for unipotent elements is Lemma 8.3 of Borel and Tits - Groupes réductifs (MSN), and I suppose you could apply it to the group generated by your $x$.
A: YCor's question about your definition of "nilpotent" is definitely in order, because Borel and Springer already defined this term in general for affine (=linear) algebraic groups over an arbitrary infinite field $k$ as an element lying in the Lie algebra of a unipotent subgroup of $G$ defined over $k$.     This two-part paper came out of their collaboration at the 1965 AMS summer institute in Boulder, with the longer second part here.    The first part is in the institute proceedings, published by AMS as PSPM 9.
In characteristic 0, there are some stronger results in the 1969 W.A. Benjamin notes from Borel's lectures written up by Bass (see $\S7$ of his second edition published by Springer as GTM 126) and in my Chapter V of Springer GTM 21.  Some of his discussion of "replicas" goes back to Chevalley's treatment in vol. 2 of his later abandoned series of books on Lie groups, Lie algebras, and affine algebraic groups.
Note that your own assumptions on $G$ are too strict.   For example, $G$ need not be reductive or semisimple.
