Why are the dineutron and diproton unbound?
Great question!
My answer would be that in order to get a bound state, we need to have a potential that is deeper than the kinetic energy the two particles have. We have a better chance of getting a potential of the right depth to bind two nucleons if:
(1) They don't repel charge-wise. Compared to nucleon interaction the Coulomb force isn't that strong, but it's still worth considering. This makes two protons look a little less attractive as a bound system.
(2) The two nucleons being aligned in spin gives some extra binding. This is a property of the nuclear force, in which there is a term
${S}_{12}(\hat{r},\hat{r}) = (\sigma_{1}\cdot\hat{r})(\sigma_{2}\cdot\hat{r})-3(\sigma_{1}\cdot\sigma_{2})$.
The two nucleons in a zero orbital angular momentum state (state of lowest energy) can only align in spin if they are antialigned in isospin, by Pauli exclusion.
The second point is the most important one: nucleons antialigned in isospin can be aligned in spin in an orbital angular-momentum zero state (S-state), by Pauli. This alignment of spins gives them the extra binding they need to form a bound state, because of the dot-product in spin in in the NN interaction, as in the term I mentioned. A proton and a neutron are antialigned in isospin. This means they can align in spin in an S-state, which gives them the "extra" bit of binding energy and lets them stay bound.
The nucleon-nucleon interaction has a short range, roughly 1 fm. Therefore if there were to be a bound dineutron, the neutrons would have to be confined within a space roughly this big. The Heisenberg uncertainty principle then dictates a minimum uncertainty in their momentum. This amount of momentum is at the edge of what theoretical calculations suggest the strong nuclear force could successfully fight against. Experiments in 2012 give evidence that the dineutron may be weakly bound, or that it may be a resonance state that is close enough to bound to create the same kind of strong correlations in a detector that you would get from a dineutron. So it appears that the strong nuclear force is not quite strong enough, but this is not even clear experimentally.
If the dineutron isn't bound, the diproton is guaranteed not to be bound. The nuclear interaction is the same as in the dineutron, by isospin symmetry, but in addition there is an electrical repulsion.