Cross-posted from MSE.

Many interesting manifolds can be expressed as $ G/H $ for $ G $ a connected Lie group and $ H $ a maximal closed subgroup. Examples include the projective spaces $ \mathbb{C}P^n \cong \operatorname{SU}_n/U_{n-1} $ where $ U_{n-1} $ is maximal for $ n \geq 3 $, and $ \mathbb{R}P^n \cong \operatorname{SO}_n/O_{n-1} $, again $ O_{n-1} $ is maximal for $ n \geq 3 $. Another example is the Poincare homology sphere $ \operatorname{SO}_3(\mathbb{R})/A_5 $.

Solvmanifolds provide many interesting examples of manifolds, especially of torus bundles over tori (a solvmanifold is a manifold of the form $ G/H $ for $ G $ a solvable Lie group).

The examples I list above of manifolds $ G/H $, $ H $ maximal, all have the property that $ G $ is connected semisimple (indeed simple).

This leads me to wonder about the opposite case: maximal closed subgroups $ H $ of connected solvable Lie groups $ G $. Do they even exist?

Let $ G $ be a connected Lie group.

If $ G $ is abelian then certainly $ G $ does not have any maximal closed subgroups. Does the same hold for $ G $ solvable?

$\DeclareMathOperator\Ab{Ab}$Comment: Let $ G' $ be the commutator subgroup of the connected group $ G $. Let
$$
\Ab: G \to G/G'
$$

be the abelianization map. If $ H $ is a maximal closed subgroup of $ G $ then we must have
$$
\Ab(H)=G/G'
$$
because if $ \Ab(H) $ was properly contained then $ \Ab(H) $ would be a maximal closed subgroup of the connected abelian group $ G/G' $ which is impossible. In particular that implies that $ H $ does not contain $ G' $ (because if $ \Ab(H)=G/G' $
and $ H $ contained $ G' $ that would imply that $ H $ is all of $ G $, contradicting maximality).

Update: Recall that $ G $ is always a connected Lie group.

If $ G $ is nilpotent then there does not exist any maximal proper closed subgroup (proved in the original answer of LSpice).

If $ G $ is non-nilpotent then there does exist some maximal proper closed subgroup. We prove this with two cases.

$\DeclareMathOperator\Lie{Lie}$If the non-nilpotent group $ G $ is moreover non-solvable then we appeal to basically a Levi decomposition. $ \Lie(G) $ can be written as $$ \Lie(G)= \mathfrak{g}_\text{solv} \rtimes \mathfrak{g}_\text{ss}. $$ Let $ G_\text{solv} $ be a maximal solvable closed connected subgroup of $ G $ corresponding to the Lie subalgebra $ \mathfrak{g}_\text{solv} $. Let $ G_\text{ss} $ be a maximal semisimple closed connected subgroup of $ G $, corresponding to the Lie subalgebra $ \mathfrak{g}_\text{ss} $. Pick $ H_\text{max} $ to be a maximal proper closed subgroup of $ G_\text{ss} $ (there are lots of fairly well known maximal closed subgroups of semisimple groups). Then the group generated by $ G_\text{solv} $ and $ H_\text{max} $ should be roughly $ G_\text{solv} \rtimes H_\text{max} $ and should be a maximal proper closed subgroup of $ G $.

For the case that the non-nilpotent group $ G $ is solvable, then apply the answer from YCor (accepted below) which shows that a solvable non-nilpotent Lie group $ G $ must have a quotient which is one of the four solvable non-nilpotent subgroups of $$ \operatorname{AGL}_1(\mathbb{C}) \cong \mathbb{C}^* \ltimes \mathbb{C} $$ that YCor lists below. In that case there is a maximal proper closed subgroup of the quotient so we can pullback through the quotient map to get a maximal proper closed subgroup of the solvable non-nilpotent Lie group $ G $.

This proves the claim, from YCor's comment, that a connected Lie group $ G $ has a maximal proper closed subgroup if and only if $ G $ is non-nilpotent.