Like quark baryons

From your profile I see that you have no mathematical background in physics at the level of discussing theories, so this answer is general.

The quark model of baryons developed in order to model with mathematical functions the data of particle physics in order to have an exact tool to predict new experimental results. Successful predictions validate a model and give confidence to further theory development and further experiments.At the same time the quark model "explained" the existence of protons and neutrons, and of nuclei made up of protons and neutrons, and how the strong nuclear force acts.

Very basic in modeling are conservation laws that cannot be broken,because experimentally the laws always hold, that is why they are called laws, they are the axioms of the physics theory. Charge conservation is one of them.

If you look at the quark table ,

you see that quarks have a specific charge, and specific other quantum numbers that have to be conserved in order for a reaction to happen, or a bound state to be allowed.

As all macroscopic matter is neutral, and the charge of proton is positive and neutron neutral, there are only unique combinations that can make up the macroscopic world, and that is what the theory that developed the quarks makes sure of. There are no contradictions, no stable baryons other than the proton and neutron, by construction in order to fit the observable world.

As the comments tell you there are combinations of quarks stable enough through the attraction of the strong force, and allowd by the group theory mathematics of the model that have been observed in particle physics experiments, and more are predicted and experiments are looking for them.

Does there have to be 2 different types of quarks for the strong force to hold them together

The quantum numbers have to sum up correctly for a bunch of quarks to make a resonance, a short lived elementary quantum entity. You have to go into the details of the mathematics .

or do they just decay into something else?

such combinations, if allowed by quantum number conservation, have a very short lifetime, because of the strong interaction, because there are always lower energy states to decay to

There are particles with three up quarks and three down quarks i.e. the $Delta^{++}$ and $Delta^-$ particles. These are two of the four particles in the Delta baryon family.

In principle there is no restriction to having three of the same flavour quarks in a baryon - for example the $Omega^-$ baryon has three strange quarks. However no such particles are known for the heavy quarks, charm, bottom and top. Any such baryons would be very massive and very short lived. Indeed no baryons are known containing even one of the heaviest quark, the top quark, because the top quark decays too quickly to form baryons.

Back in the days of the particle zoo, when our baryon count went from proton and neutron to, well, the zoo, people started asking the exact same question: "Why do these particles exist?", and, "Why don't those particles exist?"

The answer is symmetry, and the exact nature of that symmetry led to the quark model and the 1969 Nobel Prize in physics.

The driving factor is the Pauli Exclusion Principle, or, more correctly, the quantum field theory version of it, which is called the Spin-Statistics theorem: fermion states need to change sign if you exchange any pair of particles.

There is little to offer in the realm of intuition as to why that is true. So it's best to use an example.

Consider the deuteron, the simplest nucleus. Here one considers the neutron and proton as identical particles, with a spin-like quantum number (called iso-spin), where the proton is "up", and the neutron is "down". They appear in the deuteron as:

$$ |drangle = frac 1 {sqrt 2}(|pnrangle - |nprangle)$$

Here: $|pnrangle$ means the 1st particle is a proton and the 2nd particle is a neutron, and likewise for $|nprangle$.

If you interchange them:

$$ |drangle rightarrow frac 1 {sqrt 2}(|nprangle - |pnrangle) = -frac 1 {sqrt 2}(|pnrangle - |nprangle) = -|drangle$$

The sign of $ |drangle$ changes. That's what antisymmetric under interchange means. (Note also: in deuteron, a single nucleon doesn't have a definite particle identity. It is 50% proton and 50% neutron).

All of these ideas carry over to the up and down quarks of the quark model.

When applied to baryons, it means the total wave function:

$$ psi = psi_{rm space}psi_{rm color}psi_{rm spin}psi_{rm flavor} $$

has to be antisymmetric.

Protons and neutrons are spherical, so $psi_{rm space}$ is symmetric (the details follow from the quantum treatment of orbital angular momentum). $psi_{rm color}$ is antisymmetric (which is a phenomenological rule of QCD). Thus, the product $psi_{rm spin}psi_{rm flavor}$ must be symmetric.

How that works for the proton is shown here: Proton spin/flavor wavefunction . (3-particle symmetries in product spaces is involved, but straightforward).

What is germane to your question is this: consider 3 up quarks:

$$ psi_{rm flavor} = |uuurangle $$

If I exchange any two particles in the flavor state, I get:

$$ psi_{rm flavor} rightarrow |uuurangle = +psi_{rm flavor} $$

It is symmetric. Hence, the spin wave function must also be symmetric, and when you combined three spin 1/2, the rules of quantum spin tell you the state must be spin 3/2 state, e.g.:

$$ psi_{rm spin} = |uparrowuparrowuparrowrangle $$

So that's it: protons and neutrons are spin 1/2. The symmetry requirements rule out at spin 1/2 stable baryon with 3 up quarks.

You can circumvent this with orbital angular momentum. An $L=1$ state makes $psi_{rm space}$ antisymmetric. I think this state is the $N^*(1520)$ listed here: https://pdg.lbl.gov/2019/tables/rpp2019-tab-baryons-N.pdf .

The more well known baryons are the spin-3/2 $Sigma$ baryons (https://pdg.lbl.gov/2019/tables/rpp2019-tab-baryons-Sigma.pdf), which have a symmetric spin state and a symmetric flavor state.