You have many misconceptions and are failing to see the rigour that is already being presented. This answer will even just cite A&M, because they have commented upon basically everything that you are having misconceptions on. The equations I am using, if they are labelled in the style of A&M, appear inside A&M.
Impossibility of $\mathbb R^3$ $\vec k$-space
In problem 1 of Chapter 8, A&M covers a direct counterexample. Figure 8.6 plots out Equation (8.76) and shows that for $\mathbb R$ $k$-space in a strong potential, there are ranges of $k$ where there is no possible solution. Note that Figure 8.6 explicitly plots $k$ on the y-axis as only available in 1BZ, whereas the horizontal x-axis is plotting $K$ that is really the energy eigenvalue. I take this as that it is futile to insist that there would be a bijection between $\mathbb R^3$ $\vec k$-space and the joint eigenbasis, because a too strong (or too weak) potential can kill some previously expected states.
Authorial Intent for $n^\text{th}$ Band Labelling
$$\tag{8.6}\text{You have quoted}\qquad\psi(\vec r+\vec R)=e^{i\vec k\cdot\vec R}\psi(\vec r)$$
But this comes by you ignoring
\begin{align}
\tag{8.3}\psi_{n\vec k}(\vec r)&=e^{i\vec k\cdot\vec r}u_{n\vec k}(\vec r)\\
\tag{8.4}u_{n\vec k}(\vec r+\vec R)&=u_{n\vec k}(\vec r)\\
\tag{8.5}\psi_{n\vec k}(\vec r+\vec R)&=e^{i\vec k\cdot\vec R}u_{n\vec k}(\vec r)
\end{align}
and the footnotes 2 and 3 on that page. You should also look at the General Remarks About Bloch's Theorem, points 2, 3 and 4, which explicitly covers why there is a need for band indexing and where $\vec k$ is meant to be considered from.
I am very worried about you misunderstanding what I am going to write next, so I am going to break with A&M notation for a while. The reciprocal lattice vector $\vec G=n_1\vec b_1+n_2\vec b_2+n_3\vec b_3$ with $(n_1,n_2,n_3)\in\mathbb Z^3$ is labelled as $\vec K$ in A&M, which is very easy to confuse with $\vec k\in$1BZ i.e. strictly fractional coefficients (real-valued in the continuum limit) rather than integer coefficients. If you add any $\vec G$ to $\vec k$, bringing them out of 1BZ, then in Equation (8.5) in particular, it is very clear that the $e^{i\vec G\cdot\vec R}=1$ by definition. This is thus a redundant description and we do need to limit ourselves to 1BZ if we want to avoid double-counting, when we use the band indexing.
Information from Second Proof
\begin{align}
\tag{8.1 }U(\vec r)&=U(\vec r+\vec R)\\
\tag{8.31}\text{directly implies}\qquad U(\vec r)&=\sum_{\vec G}U_{\vec G}\,e^{i\vec G\cdot\vec r}
\end{align}
Similarly,
$$\tag{8.43&44}\psi_{n\vec k}(\vec r)=e^{i\vec k\cdot\vec r}u_{n\vec k}(\vec r)\qquad\bigwedge\qquad u_{n\vec k}=\sum_{\vec G}c_{n,\vec k-\vec G}\,e^{-i\vec G\cdot\vec r}$$
openly tells you that it is actually that all $\vec G$ needs to contribute to each Bloch state. You really cannot distinguish which state is actually which, between the $\vec k$ and $\vec k+\vec G$, because all of these come together to make one single Bloch state.
Extended Zone Scheme
If you look at Figure 9.4, A&M explicitly constructs the extended zone scheme, the reduced zone scheme, and the periodic zone scheme. The periodic zone scheme completely does away with even attempting to avoid double-counting. The reduced zone scheme, that all of us are telling you about, is the least confusing in practice, because the 1BZ is completely unambiguous, and you can just use the band index to go up to the energy level you are interested in.
The extended zone scheme is doable in 1D, but if you have more dimensions, it is exceedingly easy to miss or double-count. It might not even be interpretable, and the ease at which it looks in Figure 9.4 is because it is the most simple, 1D weak potential limit. Only in the case that you can guarantee that you have established a bijection from the reduced zone scheme to the extended zone scheme, then, and only then, might you say that you can drop the band index and use the extended zone scheme as the proxy for band indexing. Which, again, might not exist in strong potentials.
I hope that from here onwards, it is clear that Bloch's theorem is actually much more natural and obvious in the reduced zone scheme than in the extended zone scheme that you are looking at. Basically every author is going to want to start with a treatment in reduced zone scheme, because of how the explicit 2nd proof goes. The sum on reciprocal lattice vectors is extremely telling.
But you are also having a misconception of what it is that the argument actually is about. To illustrate that, I have to pick a detour to another system, where we can discuss the details with less confusion.
The most familiar example to try is the Hydrogen atom. The task is to get a basis for the Hilbert space. To do that, we seek a maximally commuting set of operators with the Hamiltonian. For the Schrödinger or Pauli cases, this set will be
$\{\hat{\mathcal H},\hat{L^2},\hat{S^2},\hat{J^2},\hat{J_z}\}$ with the principal quantum numbers $(n,\ell,\frac12,j,m_j)$ to go with them. The equivalents in the Dirac case are $\{\hat{\mathcal H},\hat{J^2},\hat{J_z},\hat K\}$ and $(n,j,m_j,\kappa)$. In either case, spherical symmetry means that the energy eigenvalues can only depend upon $n$ and $j$, never $m_j,\ell$ or $\kappa$, at least until hyperfine corrections and above forces us to do even more work.
There is no way to do an extended zone scheme here. In order to express the full energy eigenvalues, the quantum number $n$ is a necessity. It is thus rather weird that you are insisting to do away with it in the case of Bloch's theorem. After all, Figure 9.4 was assuming that there was no lattice at all, just a weird periodic weak potential with one electron in it. Then, and only then, can it have the quadratic that looks so beautiful. In reality, each energy level $n$ of an atom, say, will spawn at least one band. It will be horrendous.
Even if we do not go that far, note that it is important to have a maximally commuting set. The consequence of not having a maximally commuting set, is that there will be a failure to distinguish degenerate states. That is, if you only have $\{\hat{\mathcal H},\hat{J^2}\}$, then their joint eigenbasis spans the entire Hilbert space with no problems, but then the degenerate subspaces will be confusing messes. There is never any worry that the eigenbasis will be insufficient—by postulation, any single Hermitian (observable) operator has an eigenbasis that spans the entire Hilbert space. The only problem is that it is possible for the degeneracies to cause confusion.
The relevance of the above consideration to the problem at hand, is that we are NEVER trying to establish a ``bijection" or whatnot. That is simply not the correct logic. The available state space is also not just something you can insist to throw all possible values into. For example, if you choose $\ell>j$, then there will be no possible state. Similarly, if you pick $n=-0.2$, then obviously there is no state, even if $\vec k\in$1BZ. That is, nobody proving Bloch theorem should be burdened with stating that not all combinations of the stuff is possible. That $\vec k+\vec G$ is 1) conditionally well-defined in the extended zone scheme by replacing band indexing or 2) illegal in the reduced zone scheme or 3) double-counted in periodic zone scheme, is actually outside the scope of Bloch's theorem.
In fact, Bloch theorem explicitly does not cover spin, and so it is manifestly not a maximally commuting set of operators with the Hamiltonian.