Here is another answer in case $G = \mathrm{PGL}_2 \times \mathrm{PGL}_2$, just using elementary calculations. So let $H \subset G$ be a connected maximal algebraic subgroup. For $i = 1, 2$, denote by $p_i \colon G \to \mathrm{PGL}_2$ the projection to the $i$-th factor.

If $p_1(H) \neq \mathrm{PGL}_2$, then $p_1(H)$ is a maximal (connected) subgroup of $\mathrm{PGL}_2$ and thus $H = B \times \mathrm{PGL}_2$ for
some Borel subgroup $B$ of $\mathrm{PGL}_2$. Analogously, if $p_2(H) \neq \mathrm{PGL}_2$, then $H = \mathrm{PGL}_2 \times B$.

Now, assume $p_1(H) = p_2(H) = \mathrm{PGL}_2$. Let
$N \subseteq H$ be a normal subgroup of $H$. Since $\mathrm{PGL}_2$ is simple, $p_i(N) = 1$ for $i =1, 2$ and hence $N$ lies in the kernel of $p_1$ and of $p_2$. Thus $N = 1$ and therefore $H$ is simple. Since $H$
is a subgroup of $\mathrm{PGL}_2 \times \mathrm{PGL}_2$, it must be isomorphic to $\mathrm{PGL}_2$. Thus, $p_i \colon H \to \mathrm{PGL}_2$ is an isomorphism for $i=1, 2$ and hence H is conjugate to the diagonal embedding (here one uses the fact that all automorphisms of $\mathrm{PGL}_2$ are inner).

[ADDED] There is a more direct argument in case $p_1(H) = p_2(H) = \mathrm{PGL}_2$: the kernel of $p_1 |_H \colon H \to \mathrm{PGL}_2$ is the normal subgroup
$H \cap \{ 1 \} \times \mathrm{PGL}_2$ of $H$. Since
$p_2(H) = \mathrm{PGL}_2$, one can easily see that
$H \cap \{ 1 \} \times \mathrm{PGL}_2$ is normal in $\{ 1 \} \times \mathrm{PGL}_2$.
Since $H \subsetneq \mathrm{PGL}_2 \times \mathrm{PGL}_2$, this implies that $H \cap \{ 1 \} \times \mathrm{PGL}_2$ is trivial and thus $p_1 |_H$ is an isomorphism. Analogously one sees that
$p_2 |_H$ is an isomorphism. Hence $H$ is conjugate to the diagonal embedding (here one uses the fact that all automorphisms of $\mathrm{PGL}_2$ are inner).

[ERRATUM] The direct argument above only shows that $p_1 |_H$ and $p_2 |_H$
are bijective.