A cycle of $a_i$ for $k$ odd steps and $n$ even steps must satisfy a product formula
$$ 2^n = (3+{1\over a_1})(3+{1\over a_2})\cdots (3+{1\over a_k})
$$
Because the rhs is larger than $3^k$ we must have that $n \ge \lceil k \cdot \ln_2(3) \rceil \approx \lceil 1.5 k \rceil $
If we want $n=2k$ then we have
$$ 2^{2k} = (3+{1\over a_1})(3+{1\over a_2})\cdots (3+{1\over a_k}) \\
4^{k} = \underset{ k \text{ - parentheses}}{\underbrace{(3+{1\over a_1})(3+{1\over a_2})\cdots (3+{1\over a_k})}} \\$$
and from this, since no parenthese on the rhs can be larger than $4$, no other one can be smaller than $4$ and thus all of them must equal $4$ and finally all $a_i=1$ is required.
So this all defines the "trivial cycle" $1 \to 1 \to \cdots $ of length $k$ and no other solutions are possible.
I have a bit of discussion in my
text on my homepage which might interest you although this is no (serious) "reference".
(Unfortunately I do not know any reference which discusses this detail explicitely but of course the product formula and some consequences of it are well known at least since R.Crandall1 in the 70'ies)
1 Crandall, R. E., On the ”3x+1” problem, Math. Comput. 32, 1281-1292 (1978). ZBL0395.10013.
For some integers $z_i$ and $a_i$, we have
$$x_i \equiv z_i \pmod{2^{k_i}} \implies x_i = 2^{k_i}a_i + z_i \tag{1}\label{eq1A}$$
Next, there are $k_i$ odd integer results in a row after repeated applications of the $T$ function starting with $x_i$. This gives for the first one,
$$\begin{equation}\begin{aligned}
T^{1}(x_i) & = \frac{3x_i + 1}{2} \\
& = \frac{3(2^{k_i}a_i + z_i) + 1}{2} \\
& = \frac{3(2^{k_i}a_i) + 3(z_i) + 1}{2}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
The next one becomes
$$\begin{equation}\begin{aligned}
T^{2}(x_i) & = \frac{3T^{1}(x_i) + 1}{2} \\
& = \frac{3\left(\frac{3(2^{k_i}a_i) + 3(z_i) + 1}{2}\right) + 1}{2} \\
& = \frac{\frac{3^2(2^{k_i}a_i) + 3^2(z_i) + 3}{2} + \frac{2}{2}}{2} \\
& = \frac{3^2(2^{k_i}a_i) + 3^2(z_i) + 3 + 2}{2^2}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
The third one is
$$\begin{equation}\begin{aligned}
T^{3}(x_i) & = \frac{3T^{2}(x_i) + 1}{2} \\
& = \frac{3\left(\frac{3^2(2^{k_i}a_i) + 3^2(z_i) + 3 + 2}{2^2}\right) + 1}{2} \\
& = \frac{\frac{3^3(2^{k_i}a_i) + 3^3(z_i) + 3^2 + 3(2)}{2^2} + \frac{2^2}{2^2}}{2} \\
& = \frac{3^3(2^{k_i}a_i) + 3^3(z_i) + 3^2 + 3(2) + 2^2}{2^3}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
By continuing this, the general result for $T^{q}(x_i)$ for any $1 \le q \le k_i$, which you can fairly easily prove by induction & which I'll leave to you to do, becomes
$$\begin{equation}\begin{aligned}
T^{q}(x_i) & = \frac{3^{q}(2^{k_i}a_i) + 3^{q}(z_i) + \sum_{j=0}^{q-1}3^{q-1-j}2^{j}}{2^{q}} \\
& = \frac{3^{q}(2^{k_i}a_i) + 3^{q}(z_i) + 3^{q-1}\sum_{j=0}^{q-1}3^{-j}2^{j}}{2^{q}} \\
& = \frac{3^{q}(2^{k_i}a_i) + 3^{q}(z_i) + 3^{q-1}\sum_{j=0}^{q-1}\left(\frac{2}{3}\right)^{j}}{2^{q}} \\
& = \frac{3^{q}(2^{k_i}a_i) + 3^{q}(z_i) + 3^{q-1}\left(\frac{1-\left(\frac{2}{3}\right)^{q}}{1-\frac{2}{3}}\right)}{2^{q}} \\
& = \frac{3^{q}(2^{k_i}a_i) + 3^{q}(z_i) + 3^{q}\left(\frac{3^{q} - 2^{q}}{3^{q}}\right)}{2^{q}} \\
& = \frac{3^{q}(2^{k_i}a_i) + 3^{q}(z_i + 1) - 2^{q}}{2^{q}} \\
& = 3^{k_i}\left(2^{k_i-q}\right)a_i + \frac{3^{k_i}(z_i + 1)}{2^{q}} - 1
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
With $q = k_i$, \eqref{eq5A} becomes
$$T^{k_i}(x_i) = 3^{k_i}a_i + \frac{3^{k_i}(z_i + 1)}{2^{k_i}} - 1 \tag{6}\label{eq6A}$$
For $T^{k_i}(x_i)$ to be an integer requires the middle term's numerator to be a multiple of $2^{k_i}$. Since $\gcd(3^{k_i}, 2^{k_i}) = 1$, this gives for some integer $r$ that
$$2^{k_i} \mid 3^{k_i}(z_i + 1) \implies 2^{k_i} \mid z_i + 1 \implies z_i = r\left(2^{k_i}\right) - 1 \tag{7}\label{eq7A}$$
Thus, $r = 0$ gives $z_i = -1$ to be a solution. Also, the middle term in \eqref{eq5A} becomes $0$ so the equation simplifies to $T^{q}(x_i) = 3^{k_i}\left(2^{k_i-q}\right)a_i - 1$. As such, for each $q \lt k_i$, it's an odd integer, matching the requirement that these values all be odd. In addition, \eqref{eq1A} then becomes your Observation $1$, i.e.,
$$x_i = 2^{k_i}a_i - 1 \tag{8}\label{eq8A}$$
Note with $z_i = -1$ that \eqref{eq6A} simplifies to
$$T^{k_i}(x_i) = 3^{k_i}a_i - 1 \tag{9}\label{eq9A}$$
With the definitions being used, after $k_i$ iterations of applying $T$ starting with $x_i$, the set of odd numbers end and an even number is the result at this point (note this means $a_i$ must be odd). The value increases when $T$ is applied to each odd number, but it decreases with each even number, so $T^{k_i}(x_i)$ is a local maximum, i.e., it's your $y_i$. Thus, \eqref{eq9A} gives your Observation $2$, i.e.,
$$y_i = 3^{k_i}a_i - 1 \tag{10}\label{eq10A}$$
Best Answer
Just to add some more combinatorical information.
I looked empirically for the number of possible sets of exponents $A_k$ with the restriction that $S=\lceil N \log_2(3) \rceil$ - that means the sets of orbit-candidates which must been tested for cyclicity.
Here, rotations of the exponents, for example $A,B,C,D$ and $B,C,D,A$, are taken as duplicate list entries of a cycle-candidate and are only inserted as one instance in the final list.
Here is the empirical list of sets of cycle-candidates for $N=2 .. 8$:
My q&d-routine to detect this is extremely time-consuming; but the results and very likely the continuation of this can be found in the OEIS, hidden in the following (infinite) rectangular array (headers-lines are mine, "maxA" is reference to mine, square brackets indicate my empirical numbers $c$ of-orbits-to-be-tested):
Update: The T()-formula in OEIS is extremely helpful!
Here is the list of systematic $c(N)$ (= for each $N$) with $N=1..29$: $$ [1, 2, 2, 5, 7, 22, 66, 99, 335, 504, 1768, 6310, 9690, 35530, 54484, 204347, 312455, 1193010, 4552275, 7056280, 27293640, 42181080, 165056400, 644637006, 1005633632, 3964522026, 6167026726, 24512635642, 38036848410,\ldots]$$
(Here the same for Collatz-over-the-negative-numbers, $S=\lceil N \log_2(3) \rceil -1$) $$ [1, 1, 1, 3, 3, 10, 30, 43, 143, 201, 728, 2652, 3876, 14550, 21318, 81719, 120175, 468754, 1820910, 2731365, 10752060, 16128424, 64188600, 254463276, 386782164, 1547128656, 2349343610, 9470798326, 14369476066,...]$$ $\phantom{aaaaaaaaaaaa}$ (Note that there are $3$ cycles in the negative numbers)
Your question
is surely to be answered as "yes"; the number of candidate-orbits (even if rotations are ignored) seem to be exponentially in $N$ (your $k$).
Appendix: The Pari/GP-routines I've used is: