Yes, please don't feel bad about this confusion. It is for semi-historical reasons as KReiser rightly pointed out above, and can be quite confusing.
Namely, let us fix a field $K$ and an (affine) algebraic group $G$ over $K$. Then, for any separable$\color{red}{^{(\ast)}}$ algebraic extension $L/K$ one can form the group of $L$-rational characters
$$X^\ast_L(G):=\mathrm{Hom}(G_L,\mathbb{G}_{m,L}),$$
where these are homomorphisms of group $L$-schemes. There is somewhat varying terminology in this regard, but often people call $X_{K^\mathrm{sep}}$ the group of characters of $G$, and denote it just $X^\ast(G)$ (or $X(G)$). They then might call the elements of $X_K^\ast(G)$ the rational characters of $G$. Although, again, this is author dependent, and so you have to sort of feel out the definition in each case.
To understand the relationship between the rational characters over differing fields, it's useful to introduce some extra structure. If $L/K$ is Galois, with Galois group $\Gamma_{L/K}$, then there is a natural action of $\Gamma_{L/K}$ on $X^\ast_L(G)$ given as follows:
$$\sigma\cdot \chi:= \sigma_G\circ \chi\circ\sigma_{\mathbb{G}_{m,L}}^{-1}.$$
Here for a $K$-scheme $X$ and $\sigma$ in $\Gamma_{L/K}$, I am denoting by $\sigma_X$ the natural morphism $X_L\to X_L$ which is the identity on the $X$-factor and the induced map on $\mathrm{Spec}(L)$ given by $\sigma^{-1}\colon L\to L$ (you can remove the inverse here if you are OK with it being a right, opposed to left, action). It is a good exercise to check that this is a well-defined action (e.g. that $\sigma\cdot\chi$ really can still be made sense of as an element of $X^\ast_L(G)$).
From Galois descent for affine schemes (e.g. see [Poonen,§4.4]) there is a natural identification
$$X^\ast_M(G)=X_L^\ast(G)^{\Gamma_{L/M}},$$
for any tower of extensions $L/M/K$ (with $L/K$ Galois). Here I am being somewhat imprecise and identifying $X^\ast_M(G)$ with a subset of $X_L^\ast(G)$ via the (injective) group map
$$X^\ast_M(G)\to X^\ast_L(G),\qquad \chi\mapsto \chi_L,$$
(the base change map).
In particular, one sees that the natural map
$$X_K^\ast(G)\to X^\ast_{K^\mathrm{sep}}(G)=X^\ast(G),$$
is a bijection if and only if $\Gamma_{K^\mathrm{sep}/K}$ acts trivially on $X^\ast(G)$. As you mentioned, if $G$ is a torus, then this is equivalent to $G\cong \mathbb{G}_{m,K}^n$, i.e. that it is split (and in fact, you can use closely related ideas to define the maximal split subtorus -- see [Springer, Proposition 13.2.4]).
I hope this clarifies things!
$\color{red}{(\ast):}$ This is a little too technical to mention above, but you can really have $L/K$ be any algebraic extension. But, if $L/K$ is purely inseparable then the natural map $X^\ast_K(G)\to X_L^\ast(G)$ is a bijection.
References:
[Poonen] Poonen, B., 2017. Rational points on varieties (Vol. 186). American Mathematical Soc..
[Springer] Springer, T.A. and Springer, T.A., 1998. Linear algebraic groups (Vol. 9). Boston: Birkhäuser.
Best Answer
$\newcommand{\Q}{\mathbb{Q}}$$\newcommand{\Z}{\mathbb{Z}}$$\newcommand{\Lie}{\mathrm{Lie}}$Let $\lambda\in X_\ast(T)$ be such that for all roots $\alpha\in\Phi(G,T)$ we have that $\langle \lambda,\alpha\rangle\ne 0$. This is always possible since, essentially, the set of $\lambda$ such that $\langle \lambda,\alpha\rangle=0$ for one $\alpha$ is a $\Z$-hyperplane in $X_\ast(T)$, and a union of finitely many such $\Z$-hyperplane cannot equal all of $X_\ast(T)$. Such a $\lambda$ is called regular. Let $Z_{G}(\lambda)$ be the scheme-theoretic centralizer of $\lambda$. We claim that $Z_{G}(\lambda)=T$.
Well, $Z_{G}(\lambda)$ is smooth and connected (see Theorem 4.1.7 of Conrad's Reductive Group Schemes notes) and since we have an obvious inclusion $T\subseteq Z_{G}(\lambda)$ it suffices to show that we have an equality of Lie algebras $\Lie(T)=\Lie(Z_{G}(\lambda))$. But, note that this latter Lie algebra can be described in terms of the subalgebra of $\Lie(G)$ where $\lambda$ acts trivially. Note though that for each root space $\mathfrak{g}_\alpha$ we have that $\lambda$ acts on $\mathfrak{g}_\alpha$ by $t^{\langle \lambda,\alpha\rangle}$. By assumption this is non-trivial for all $\alpha$ and thus the claim follows.
Note then that since $Z_{G}(\lambda)=T$, and thus solvabe, we have that $P_{G}(\lambda)$ is solvable since $P_{G}=Z_{G}(\lambda)\rtimes U_{G}(\lambda)$ (See loc. cit. for the definitions of these objects and this decomposition). But, we know that $P_{G}(\lambda)$ is a connected parabolic subgroup of $G$. Thus, $P_{G}(\lambda)$ being a solvable parabolic subgroup is necessarily a Borel.
EDIT: Another tongue-in-cheek answer is the following:
(note that the above doesn't depend on the choice of maximal split torus since all are conjugate by Grothendieck's theorem)
So, if $G$ is split with maximal torus $T$, then $T$ is also a maximal split torus and since $T$ is self-centralizing we deduce that $G$ is quasi-split! :)
Of course, this is very silly since proving the above statement requires a real deep-dive into the relative root theory of quasi-split groups, which is very complicated.