Whilst trying to understand the answer of Loren, I finally came up with an elementary explanation (no need for Steinberg’s result). Since it also gives another picture of the situation, I think it’s worth writing.

Let $G$ and $G^*$ be algebraic groups over $F$ as in the question. Let $K$ be a finite Galois extension of $F$ splitting those groups, and let $G_0\cong G_K\cong (G^*)_K$ be the corresponding split group over $K$. Let $\varphi \colon G_0\to (G^*)_{K}$ and $\psi \colon G_0\to (G)_{K}$ be the given isomorphisms. With those notations, $G^*$ (respectively $G$) is obtained from $G_0$ by twisting against the cocycle $A_{\sigma} = \varphi^{-1}(^{\sigma}\varphi ) $ (respectively $B_{\sigma} = \psi^{-1}(^{\sigma}\psi )$), where $\sigma \in \Gamma = $Gal$(K/F)$.

We can prove that $Z(G^*)$ and $Z(G)$ are isomorphic over $F$. Indeed, they are both $K/F$ forms of $Z(G_0)$, where the twisting has values in Aut $G_0 \cong G_0/Z(G_0)\rtimes $Out $G_0$. Hence, $Z(G^*)$ and $Z(G)$ only depends on the image of $A_{\sigma}$ and $B_{\sigma}$ under $H^1(\Gamma ,\text{Aut }G_0))\to H^1(\Gamma ,\text{Out }G_0) $. But those images are cohomologous because $G$ and $G^*$ are inner forms.

Going further, let $S$ (resp. $S^*$) be a maximal $F$-split torus in $G$ (resp. $G^*$), and let $T$ (resp. $T^*$) be a maximal $F$-torus containing it. The next claim is that $ \text{rk}_F T/Z(G)\leq \text{rk}_F T^*/Z(G^*)$. Indeed, since $G/Z(G)$ and $G^*/Z(G^*)$ are inner forms of each others, the so called $*$-action on their Dynkin diagram is the same. Hence $T/Z(G)$ cannot have more characters defined over $F$ than $T^*/Z(G^*)$. Also note that $\text{rk}_F T/Z(G) = \text{rk}_F T^*/Z(G^*)$ iff $G/Z(G)$ is quasi-split, which in turn is equivalent to $G/Z(G)\cong G^*/Z(G^*)$ by uniqueness of quasi-split forms in an inner class.

EDIT:
To prove the two last sentences of the previous paragraph, we can assume that the groups are adjoint simple. Also, since Weil restriction is encoded in the $*$-action, we have $G=R_{F'/F}(G')$ and $G^* = R_{F'/F}({G^*}')$ for some unique absolutely simple $G'$ and ${G^*}'$ that are also inner forms of each other, so we can even assume that the groups are adjoint absolutely simple.

Hence, we are looking at a single Dynkin diagram with a $*$-action. Now conlcude by observing that the rank of $G^*$ is the number of orbits of the Dynkin diagram, and that the rank of $G$ is the number of orbits such that the corresponding invariant character of $T$ restricts to a non-trivial character of $S$. In terms of Tits index, this is just saying that for a quasi-split groups, all orbits are circled, while for an inner form of it, you add some anisotropicity, so that some orbits are not circled anymore. For the relation with quasi-splitness, recall that the group is quasi-split iff all orbits of the $*$-action are circled in the index.

Finally, the inequality $\text{rk}_F G\leq \text{rk}_F G^*$ holds because given an exact sequence of diagonalizable $F$-groups $1\to A\to B\to C\to 1$, $\text{rk}_F(B) = \text{rk}_F(A) + \text{rk}_F(C)$. In case of equality, as noted before, $G/Z(G)$ is quasi-split, hence so is $G^{der}$. Thus $G^{der}\cong (G^*)^{der}$, so that $G\cong G^*$, because a connected reductive group $H$ is reconstructed from the triple $(H^{der},Z(H),Z(H^{der})\hookrightarrow Z(H))$.