First one should neglect the electromagnetic force; then - regarding the strong force only - the diproton and the dineutron are two degenrate states of isospin =+1 or -1 respectively. Both diproton and dineutron do not form a bound state!
The deuteron which has isospin = 0 (one proton and one neutron form a state I=0 as the two isospins are antiparallel) is a bound state. How can this be?
The reason is that describing the strong force via a scalar potential is a very crude approximation. Some nucleon potentials (like Woods-Saxon) that one single nucleon n=1 "feels" inside a nucleus are mean field approximations only. Such a potential is generated by averaging over all other nucleons N=2, ..., N forming the nucleus. These potentials are not sufficient to describe the nucleon-nucleon force.
Instead one has to use two-nucleon potentials which have a rather complex structure: they have several spin- and isospin-dependent parts. One isospin-neutral part is always attractive, but there is an spin-and isospin-dependent part which is attractive for the deuteron but repulsive for diproton and dineutron. The reason is that for the diproton and the dineutron (with parallel isospin) the two spins always have to be antiparallel in order to fulfill the Pauli principle; (for the deuteron both parallel and antiparallel spins would be possible in principle since the two isospins are different and therefore the Pauli principle is respected by both spin states).
So the nucleon-nucleon force is still attractive for antiparalell spins, but too weak to allow for a bound state.
Here are some notes a posted several month ago:
Looking a phenomenologically successful nucleon-nucleon potentials derived from one-pion exchange one finds
V_C(r) \sim \frac{e^{-m_\pi r}}{r}
V_S(r,S) \sim V_S(r) (\vec{s}_1\vec{s}_2)
plus further terms, e.g. tensor and LS.
For the spin-spin coupling one has
\langle \vec{s}_1 \vec{s}_2\rangle = \frac{1}{2}\left[S(S+1) - s_1(s_1+1) -s_2(s_2+1)\right]
Therefore
\langle \vec{s}_1 \vec{s}_2\rangle_{S=0} = -\frac{3}{4}
\langle \vec{s}_1 \vec{s}_2\rangle_{S=1} = +\frac{1}{4}
As both diproton and dineutron are symmetric in isospin they must by antisymmetric in spin which corresponds to S=0. So the term
V_S(r,S=0) \sim -\frac{3}{4} V_S(r)
has a minus sign whereas
V_S(r,S=1) \sim +\frac{1}{4} V_S(r)
has a plus sign.
I do not know if for S=0 this results in a repulsive potential V_C(r) + V_S(r,S=0). I have found my old notes indicating that at very low energies the scattering phases for NN scattering indicate an attractive potential for both S=0 and S=1, but that attraction is not strong enough to form a bound state.
The reason why an attractive potential does not allow for bound state is well known from ordinary quantum mechanics. In one dimension an attractive potential always allows for at least one bound state; in three dimension this is no longer true, an attractive but very flat potential has scattering solutions only.
