What you want is a specific part of something that fits into a wider framework due to Borel--Tits. Specifically, you're looking for Theorem 4.15 and Corollary 4.16 of "Groupes réductifs", Inst. Hautes Études Sci. Publ. Math., 1965. Part of this result says that if $P \leqslant G$ is a $k$-parabolic subgroup then the $k$-Levi complements of $P$ are of the form $C_G(T)$ where $T \leqslant R(P)$, the radical of $P$, is a $k$-split maximal torus of $G$. Moreover, you have $C_G(T) = G$ if and only if $T \leqslant Z(G)$.

I'll say $L \leqslant G$ is a Levi subgroup of $G$ if it is the Levi complement of a parabolic subgroup $P \leqslant G$. Now let $S \leqslant C_G^{\circ}(g)$ be a maximal torus, which is necessarily a maximal torus of $G$. Being connected reductive if $C_G^{\circ}(g)$ is contained in a parabolic subgroup $P$ then it's contained in a Levi complement of $P$. Hence, it suffices to show that $C_G^{\circ}(g)$ is not contained in any proper Levi subgroup of $G$.

As the intersection of two Levi subgroups containing a common maximal torus is again a Levi subgroup there is a unique minimal Levi subgroup containing $C_G^{\circ}(g)$, namely the intersection of all such Levi subgroups. More explicitly, this Levi subgroup is given by $C_G(Z^{\circ}(C_G^{\circ}(g)))$. Hence, $C_G^{\circ}(g)$ is contained in no proper parabolic subgroup of $G$ if and only if $Z^{\circ}(C_G^{\circ}(g)) = Z^{\circ}(G)$.

Borel--Tits' result allows you to do the same over $k$. Namely, there's a unique minimal $k$-Levi subgroup containing $C_G^{\circ}(g)$. It's realised as $C_G(S)$ where $S \leqslant Z^{\circ}(C_G^{\circ}(g))$ is a $k$-split maximal torus of $Z^{\circ}(C_G^{\circ}(g))$. Hence, we have $C_G(S) = G$ if and only $S$ is a $k$-split maximal torus of $Z^{\circ}(G)$.

What you want is a specific part of something that fits into a wider framework due to Borel--Tits. Specifically, you're looking for Theorem 4.15 and Corollary 4.16 of "Groupes réductifs", Inst. Hautes Études Sci. Publ. Math., 1965. Part of this result says that if $P \leqslant G$ is a $k$-parabolic subgroup then the $k$-Levi complements of $P$ are of the form $C_G(T)$ where $T \leqslant R(P)$, the radical of $P$, is a $k$-split maximal torus of $R(P)$. Moreover, you have $C_G(T) = G$ if and only if $T \leqslant Z(G)$. $\space$

I'll say $L \leqslant G$ is a Levi subgroup of $G$ if it is the Levi complement of a parabolic subgroup $P \leqslant G$. Now let $S \leqslant C_G^{\circ}(g)$ be a maximal torus, which is necessarily a maximal torus of $G$. Being connected reductive if $C_G^{\circ}(g)$ is contained in a parabolic subgroup $P$ then it's contained in a Levi complement of $P$. Hence, it suffices to show that $C_G^{\circ}(g)$ is not contained in any proper Levi subgroup of $G$.

As the intersection of two Levi subgroups containing a common maximal torus is again a Levi subgroup there is a unique minimal Levi subgroup containing $C_G^{\circ}(g)$, namely the intersection of all such Levi subgroups. More explicitly, this Levi subgroup is given by $C_G(Z^{\circ}(C_G^{\circ}(g)))$. Hence, $C_G^{\circ}(g)$ is contained in no proper parabolic subgroup of $G$ if and only if $Z^{\circ}(C_G^{\circ}(g)) = Z^{\circ}(G)$.

Borel--Tits' result allows you to do the same over $k$. Namely, there's a unique minimal $k$-Levi subgroup containing $C_G^{\circ}(g)$. It's realised as $C_G(S)$ where $S \leqslant Z^{\circ}(C_G^{\circ}(g))$ is a $k$-split maximal torus of $Z^{\circ}(C_G^{\circ}(g))$. Hence, we have $C_G(S) = G$ if and only $S$ is a $k$-split maximal torus of $Z^{\circ}(G)$.