One can consult the tables of Bray-Holt-Roney Dougal to work out the subgroups of $\mathrm{PSU}_m$.
- For $m=5$, we have copies of $PSL_2(11)$ and $PSU_4(4)$.
- For $m=6$ we have $PSL_2(11)$, $PSL_3(4)$ and $PSU_4(9)$.
- For $m=7$ we have $PSU_3(9)$.
- For $m=8$ we have $PSL_3(4)$.
- For $m=9$ we have $PSL_2(19)$.
- For $m=10$ we have $PSL_2(19)$ and $PSL_3(4)$.
- For $m=11$ we have $PSL_2(23)$ and $PSU_5(4)$.
- For $m=12$ we have $PSL_2(23)$ and $PSL_2(9)$.
So far, so good. Bray-Holt-Roney Dougal stops at dimension 12. For dimensions 13 to 15 one consults tables in Anna Schroeder's PhD thesis (St Andrews, 2015) and there are examples in those dimensions. For dimensions 16 and 17 one consults the thesis of Daniel Rogers (Warwick, 2017). For 16 there is an example, but there are no interesting subgroups (i.e., finite subgroups not contained in a closed positive-dimensional subgroup) of $\mathrm{PSU}_{17}$ at all, never mind linear or unitary ones.
So it seems $m=17$ is the first case where things go wrong.
$\DeclareMathOperator\SU{SU}\DeclareMathOperator\GL{GL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\Cl{Cl}$Yes the the above is the correct list of maximal closed subgroups of $ \SU_3 $.
Antoneli, Forger, and Gaviria - Maximal Subgroups of Compact Lie Groups
classifies all maximal closed subgroups of $ \SU_n $ whose identity component is not simple (here trivial counts as simple). According to this paper, pages 1013–1018, the maximal closed subgroups of $ \SU_3 $ of this type are the normalizer of the maximal torus
$$
N(T)=S(U_1 \times U_1 \times U_1): S_3
$$
as well as
$$
S(U_2 \times U_1 )\cong U_2.
$$
The maximal closed subgroups with trivial identity component are the finite groups:
$$
3.A_6
$$
of order $ 3(360)=1080 $ (known as the Valentiner group)
and
$$
\langle \zeta_3I\rangle \times \GL_3(\mathbb{F}_2)
$$
of order $ 3(168)=504 $. Both these two groups are central extensions by $\langle\zeta_3 I\rangle$ of a finite simple group. But the first is perfect central extension, indeed a Schur cover. While the second is just a direct product. A third finite maximal closed subgroup is the complex reflection group with Shephard–Todd number 25 (see Complex reflection group) and order $ 3(216)=648 $ which happens to be a central extension again by $\langle\zeta_3 I\rangle$ of the Hessian group of order 216. (Since you are from quantum computing this last subgroup would be known to you as the (determinant-1 subgroup of the) qutrit Clifford group.)
The only maximal closed subgroup of $ \SU_3 $ with nontrivial simple identity component is the direct product
$$
\langle\zeta_3I\rangle \times \SO_3(\mathbb{R}).
$$
To summarize, the maximal closed subgroups of $ \SU_3 $ are
\begin{align*}
&N(T)\\
& S(U_2\times U_1)\cong U_2\\
& \langle\zeta_3 I\rangle \times \SO_3(\mathbb{R}) \\
& \langle\zeta_3 I\rangle \times \GL_3(\mathbb{F}_2) = 3 \times PerfectGroup(168) \\
& 3.A_6=PerfectGroup(1080)\\
& S(\Cl_1(3))=SmallGroup(648,532)
\end{align*}
where $ S(\Cl_1(3)) $ is the determinant-1 subgroup of the single qutrit Clifford group. Every closed subgroup of $ \SU_3 $ is contained in one of these $ 6 $ maximal groups.
Also since unitary $ t $ designs are popular in quantum computing it may be of interest to you that $ 3.A_6 $ is a unitary 3-design and $ \zeta_3 \times \GL_3(\mathbb{F}_2) $ and $ S(\Cl_1(3)) $ are both unitary 2-designs.
This is consistent with claim 3 of my answer to Finite maximal closed subgroup of connected Lie group
that all maximal $ 2 $-design subgroups of $ \SU_n $ (all $ 3 $ designs are also $ 2 $ designs) are finite maximal closed subgroups of $ \SU_n $.
The only other closed subgroups of $ \SU_3 $ which are unitary $ t $-designs for $ t \geq 2 $ are the $ \GL_3(\mathbb{F}_2) $ subgroup of $ \langle\zeta_3 I\rangle \times \GL_3(\mathbb{F}_2) $ and the commutator subgroup of the qutrit Clifford group, SmallGroup(216,88), which has size $ 3(72)=216 $. These are again unitary $ t $-designs for $ t=2 $.
In addition to all 5 of these 2-designs there is one other Lie primitive subgroup (i.e. not contained in any proper positive dimensional closed subgroup): it is a subgroup of the qutrit Clifford group of size $ 6(36)=216 $.
Best Answer
$\newcommand{\G}{\mathcal{G}} \newcommand{\K}{\mathcal{K}} $Question: When does $ \G $ admit a finite maximal closed subgroup?
Answer : Must be one of the following two cases
From now on I will confine myself to the case that $ \G $ is connected.
In other words I will consider the statement "A connected Lie group $ \G $ has a finite maximal closed subgroup $ G $ if and only if $ \G $ is compact and simple."
The first implication is true.
Claim 1: If a connected Lie group $ \G $ has a finite maximal closed subgroup $ G $ then $ \G $ must be compact and simple.
Proof: Let $ \G $ be a connected Lie group and $ G $ a finite maximal closed subgroup. Since $ G $ is finite then $ G $ is a compact subgroup of $ \G $ so must be contained in a maximal compact subgroup, call it $ \K $. But $ G $ is a maximal closed subgroup thus we must have that $ \K=\G $ (note that $ \K $ cannot equal $ G $ since $ \K $ is connected (the maximal compact of a connected group is always connected)). So $ \G $ must be compact. If $ \G $ is not simple then there exists some morphism $$ \pi: \G \to \G_i $$ with positive dimensional kernel (here $ \G_i $ is basically one of the semisimple factors of $ \G $). Then $$ \pi^{-1}(\pi(G)) $$ is a closed positive dimensional subgroup containing $ G $, contradicting the fact that $ G $ is a finite maximal closed subgroup. Thus if a connected Lie group $ \G $ has a finite maximal closed subgroup then we can conclude that $ \G $ is simple.
However the reverse implication does not hold: $ SU_{15} $ is an example of a compact connected simple Lie group with no finite maximal closed subgroups.
To see why this is the case it is important to note that
Claim 2: For a compact connected simple Lie group $ \G $, $ G $ is a finite maximal closed subgroup of $ \G $ if and only if $ G $ is Ad-irreducible and $ G $ is a maximal finite subgroup of $ \G $.
this follows from Corollary 3.5 of Sawicki and Karnas - Universality of single qudit gates.
Since a finite subgroup of $ SU_n $ is Ad-irreducible if and only if it is a unitary 2-design we have
Claim 3: $ G $ is a finite maximal closed subgroup of $ SU_n $ if and only if $ G $ is a maximal unitary 2-group in $ SU_n $.
By inspecting Theorem 3 of Bannai, Navarro, Rizo, and Pham Huu Tiep - Unitary $t$-groups one immediately determines that $ SU_{15} $ has no finite maximal closed subgroups.
Some of the main examples of finite maximal closed subgroups of $ SU_n $ include the normalizer in $ SU_{p^n} $ of an extra-special group $ p^{2n+1} $. Here $ p $ is an odd prime. There is also a similar construction $ p=2 $. These are known as (complex) Clifford groups. Then there are infinite families of examples relating to the Weil module for $ \operatorname{PSp}_{2n}(3) $ and another family related to $ U_n(2) $. Plus many exceptional cases.
A similar normalizer construction to the above gives finite maximal closed subgroups of all the $ \operatorname{SO}(2^n) $ as normalizers of an extra-special group $ 2^{2n+1} $. This is known as the real Clifford group. For details about real and complex Clifford groups see Nebe, Rains, and Sloane - Self-Dual Codes and Invariant Theory.