For the general focus of the question: "*how many* invariant sets" I don't remember any article dealing explicitely with this. Surely my readings are incomplete, but I also don't think that there is something mentioned in Lagarias' big reference/review list to that specific question.

But let me mention some own thoughts for a more general view on cycles in Collatz-type problems.

If we use the "syracuse"-notation for the Collatz problem (and similars) then the sequence of $3a_k+1$ steps followed by $2^{A_k}$ steps can economically be encoded by a notation $a_N = T(a_1; [A_1,A_2,...,A_N])$ where we denote the number of odd steps as $N$ and the number of even steps $S$ (equalling the sum of all exponents $A_k$). The expression $T(a;[A_1,...,A_N])$ has a numerator with a part free of $a_1$ which we denote as
$$Q([A_1,A_2,...,A_N])=3^{N-1} + 3^{N-2}2^{A_1} + 3^{N-3}2^{A_1+A_2} + ... + 2^{S-A_N} \tag 1$$
Let us now write shorter $E_j(N,S)$ for some vector of exponents $[A_1,A_2,...,A_N]$.

Then the whole expression becomes
$$ a_n = T(a_1;E_j(N,S)) = {3^N\over2^S}a_1 + {Q(E_j(N,S))\over2^S} \tag 2$$
If we look at a *given vector* of exponents $E_j(N,S)$ as the independent part and the elements of the orbit $a_k$ as dependents, then we can determine a cycle by $a_n=a_1$ and
$$ a_1 = {Q(E_j(N,S)) \over 2^S - 3^N} \tag 3 $$
If we demand that $a_1$ must be positive integer then we know, that for $N=1$ and $S=2$ there is the solution of the (trivial) cycle - which very likely is the only cycle.

However, in generalization, if we allow negative and rational $a_k$ then of course we have *infinitely* many solutions or infinitely many invariant sets, namely for each combination of $N$ and $S \ge N$ a collection of invariant set (depending of the composition of $E_j(N,S)$ by a vector of $A_k$). Trivially, if we ask for the generalized Collatz problem $3x+r$ then for $r=2^S-3^N$ we have a solution in the integers, so the set of $3x+r$-problems cover the set of integer solutions
$$ a_1 = r \cdot {Q(E_j(N,S)) \over 2^S - 3^N} \tag 4$$
and thus the (infinite) set of invariant sets analoguously to the generalized Collatz-problem with $ a_k \in \mathbb Q$

The same is of course true for the second generalization of the Collatz problem to $m x +r$ : using $m x + 1$ and allowing $a_k \in \mathbb Q$ we find infinitely many "invariant sets" which can be recovered in $m x + r$ - problems with $r = 2^S - m^N$

However, the problem *how many* invariant sets for $m x +1$ for $a_k \in \mathbb Z$ I have not seen been handled in literature. My own heuristic is relatively short; I've recently posted the following table:
$$ \small \begin{array}{rr|r|l}\\
\text{base } m & & n \text{of cycles} & \text{transformations}& E(N,S)\\ \hline
m=2^k-1 & 3 & 4 & [1;1;...] & [2,...] \\
&&& [-1;-1;...] & [1,...]\\
&&& [-5,-7;-5,...] & [1,2,...]\\
&&& [-17,-25,-37,-55,-41,-61,-19;-17,...] & [1,1,1,2,1,1,4,...] \\ \hline
& 7 & 1 & [1;1;...] & [3,...] \\
& 15 & 1 & [1;1;...] & [4,...] \\
& 31 & 1 & [1;1;...] & [5,...] \\
& \vdots \\
& 16383 & 1 & [1;1;...] & [14,...]\\
& \vdots \\ \hline \phantom{asas} \\ \hline
m=2^k+1 & 3 & 4 & \text{see above} \\ \hline
& 5 & 4 & [1,3;1,...]& [1,4,...] \\
&&& [13,33,83;13,...] & [1,1,5,...]\\
&&& [17,43,27;17,...] & [1,3,3,...] \\
&&& [-1;-1;...] & [2,...]\\ \hline
& 9 & 1 & [-1;-1;...] & [3,...] \\
& 17 & 1 & [-1;-1;...]& [4,...] \\
& 33 & 1 & [-1;-1;...]& [5,...] \\
& \vdots \\
& 16385 & 1 & [-1;-1;...] & [14,...]\\
& \vdots \\ \hline \phantom{aaaa} \\ \hline
\text{other }m & 181 & 2 & [27,611;27,...] & [3,12,...]\\
& & & [35,99;35,...]& [6,9,...]
\end{array} \tag 5$$
The first cycles for $m=5$ and for $m=181$ are already mentioned in Crandall's 1978 paper(and also the relation between the $a_k \in \mathbb Q$ of the $3x+1$ problem and the $a_k \in \mathbb N$ in the $3x+r$ problems) - but not in the sense of a quantification of the *number of* cycles for some of that dynamical systems. The "invariant sets" for $m=3$ and $m=5$ are also in wikipedia.

So what I've found for the *integer solutions* of the $m x +1$ are either $1$, $2$ or $4$ "invariants sets" (by that small heuristic), and only for three bases $m$ exactly two or four such sets.

Of course, this is not literature and not published - and likely needs more workout in the sense and in the focus of your question.

I've tested $m<20 000$, $|a_1| < 1000$ and $N \le 30$

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