A pair of electrons has spin 0 which makes any such system a boson rather than a fermion. The Pauli exclusion principle does not apply therefore to paired electrons and any such two electrons can coexist in the same quantum state. Wouldn't it make electron shells collapse all into the lowest 1s²?
How can paired electrons still be considered fermions if their total spin is not half-integral.
Best Answer
The abstraction of two fermions as one single particle creates not an ordinary boson, but a hard core boson. Hard core bosons share the Pauli exclusion principle with fermions in the sense that two hard core bosons cannot occupy the same state. This is true regardless of wether the fermions have actually paired through an interaction, or you just arbitrarily chose to describe them in pairs
That way, even if you were to describe the electrons in an atom by pairing them into bosons, these would be hard core bosons and so they could not all collapse to the lowest orbital because that would be already occupied.
The two pictures are thus consistent, as they should be.
EDIT: I have recently come across this answer by @Chiral Anomaly to a related question. Because of that I've come to realize my answer above is incomplete, if not plainly wrong. Consequently, my arguments in the comment section below might be wrong, I will need to review them.
As they discuss in the linked answer, you actually can put two composite bosons in the same state (meaning $(b^\dagger)^2\neq 0$), as long as the bosons are formed by entangled fermions.
However, as far as I know, composite fermions, coming from entangled fermions or not, still have distinct features from pure bosons. Let us take the example from the linked post: \begin{equation} b^\dagger(f)=\sum_{n,m}f(n,m)a_n^\dagger a_m^\dagger, \tag{1} \end{equation} where $a_n^\dagger$ are orthogonal fermionic modes. First, it can be checked that their commutation relations are in general not strictly bosonic, meaning \begin{equation} [b,b^\dagger] \neq 1, \tag{2} \end{equation} for any normalization.
Moreover, if the entanglement between the generating fermions occurs inside a finite subspace (the sum over $n,~m$ has a finite number of elements $N$), then there is a maximum number of composite bosons that can be put in a given state, meaning $(b^\dagger)^{N+1} = 0$.