I tried to prove it for the shortcut version myself, please let me know if there are any mistakes
Let $E(n) := \frac{1}{2}n$, referred to as even step, and $O(n) := \frac{1}{2}(3n + 1)$, referred to as odd step. Then the Collatz function $T:\mathbb{Z} \rightarrow \mathbb{Z}$ is given by
$$
T(n) :=
\begin{cases}
E(n), & \text{if $n$ is even} \\[2ex]
O(n), & \text{if $n$ is odd}
\end{cases}
$$
This function is applied recursively by the notation
\begin{align*}
T^0(n) &:= n \\
T^k(n) &:= T(T^{k - 1}(n)), k \in \mathbb{Z}^+
\end{align*}
Let $\delta(n)$ (dropping time of $n$) be the smallest integer $k$ such that $T^k(n) < n$.
Every integer follows a certain path of applying the even and odd step functions until it reaches a smaller value, for example, $3$ follows the path $3 \xrightarrow{\text{O}} 5 \xrightarrow{\text{O}} 8 \xrightarrow{\text{E}} 4 \xrightarrow{\text{E}} 2$, so $E(E(O(O(3))))$ is the function that joins all steps in the path it took. Let $L_k$ be the function that takes all steps that $k$ made until it dropped and join into a single function, so for $k = 3$, $L_3(n) = E(E(O(O(n)))) = \frac{1}{2}(\frac{1}{2}(\frac{1}{2}(3(\frac{1}{2}(3(n) + 1)) + 1))) = \frac{9}{16}n + \frac{5}{16}$.
Let $|L_k|$ be the total amount of steps that were joined, and $P(L_k)$ be the amount of odd steps.
Lemma 1. If n is positve and $\delta(n)$ is finite, then $L_k(n) = \frac{3^{P(L_k)}}{2^{|L_k|}}n + c$, where $c$ is some non-negative constant.
Proof. The even step function $\frac{1}{2}n$ can be written as $\frac{3^0}{2^1}n$ and the odd step function $\frac{1}{2}(3n + 1)$ can be written as $\frac{3^1}{2^1}n + \frac{1}{2}$, on both steps the exponent of $2$ in the denominator is incremented by $1$, so the denominator of the coefficient will be $2^{|L_k|}$, and on the the exponent of $3$ on the numerator is incremented by $1$ on the odd step, and by $0$ on the even step, so the numerator will be $3^{P(L_k)}$. And for the constant $c$, the only things that can affect it is the addition of $\frac{1}{2}$ on the odd step and the division by $2$ on the even step, which cannot make it negative.
Lemma 2. The smallest integer $k$ such that $2^k > 3^n$ is $\lfloor n \log_2(3) \rfloor + 1$.
Proof. $2^k > 3^n \implies k > \log_2(3^n) \implies k > n\log_2(3)$. The smallest integer $k$ that satisfies $k > n\log_2(3)$ is $\lfloor n\log_2(3) \rfloor + 1$, this is because $\lfloor n\log_2(3) \rfloor$ is the biggest integer that is less than or equal to $n\log_2(3)$, so $\lfloor n\log_2(3) \rfloor + 1$ is the smallest integer that is bigger than $n\log_2(3)$. Therefore, $\lfloor n\log_2(3) \rfloor + 1$ is the smallest integer solution of $k$ for $2^k > 3^n$.
Theorem. If $n$ is positive and $\delta(n)$ is finite, then $\delta(n) \in \{\lfloor k \log_2(3) \rfloor + 1 \mid k \in \mathbb{N}_0\}$.
Proof. By definition, the first value smaller than $k$ that appear on the sequence of applying the Collatz function is $L_k(k)$, which, as seen in lemma 1, can be written as $\frac{3^{P(L_k)}}{2^{|L_k|}}k + c$, so
$$\frac{3^{P(L_k)}}{2^{|L_k|}}k + c < k$$
As $c$ is positive, it is correct to say that
$$\frac{3^{P(L_k)}}{2^{|L_k|}}k < \frac{3^{P(L_k)}}{2^{|L_k|}}k + c$$
And using the transitive property of inequalities, the previous statements imply
$$\frac{3^{P(L_k)}}{2^{|L_k|}}k < k$$
$k$ is positive and therefore can be divided without flipping the sign to get
$$\frac{3^{P(L_k)}}{2^{|L_k|}} < 1$$
$$\implies 3^{P(L_k)} < 2^{|L_k|}$$
$$\implies 2^{|L_k|} > 3^{P(L_k)}$$
As $\delta(n)$ is the minimum amount of steps necessary to reach a smaller value, just the smallest integer solution for $|L_k|$ should be considered, which as proven in lemma 2, is $\lfloor P(L_k) \log_2(3) \rfloor + 1$. Hence, $\delta(n) \in \{\lfloor k \log_2(3) \rfloor + 1\ \mid k \in \mathbb{N}_0\}$.
Notice that this does not prove that every number in the form $\lfloor n \log_2(3) \rfloor + 1$, $n \in \mathbb{N}_0$ can be a dropping time, just that every dropping time is in this form. For that it would need a proof that any natural number can be $P(L_k)$, which I could not think of one.
