Could this odd insight help explain part of the difficulty in proving the Collatz Conjecture?
I think -although the problem has some "fractal" or "self-similar" structure- there is (at least) one more aspect to be looked at.
Consider the related problem where you insert "$3x-1$" for "$3x+1$" . (This is also simply looking at the negative integers in the original Collatz problem).
You'll have a very similar structure, again "fractal" looking. But now you'll have (at least) 3 cycles.
So we must introduce one more property which allows us to distinguish between a "1 existent cycle" fractal (with the $3x+1$) and a "3 different cycles exist" fractal (with the $3x-1$) problem.
What I think helps here is, to look at the inverse iteration.
For the original Collatz-problem this means: starting from $x_0=1$ we can construct all positive numbers by the iterative rule $$ x_{k+1,m} = {x_k \cdot 2^m - 1 \over 3} \text{for all $m$ where this is possible} $$ (only roughly defined) so $x_0=1$ defines somehow a root from where a stem, infinitely many twigs and leaves can be constructed - and with their rises and falls reach all positive numbers.
But for the $3x-1$-problem (the Collatz on the negative integers) this means: we need 3 roots, and each of the $3$ roots constructs its own specific tree with its own infinite sets of twigs and leaves - with their own risings and falls, but not overlapping and all 3 trees are needed to cover/construct all negative (in the Collatz-problem) numbers!
After that, the question of L. Collatz might be specified: "is it true, that on the positive numbers one root is sufficient, while on the negative numbers we need $3$ roots (even partially organized in cycles) to construct all (negative) numbers?"
So the "fractality" alone does not suffice; we need also an explanation for the question why the the set of positive numbers needs only one root and not -for instance- three as in the $3x-1$ problem.
As has been noted in the comments, this is called the $2$-adic valuation of $n$ and is noted as $v_2(n)$. In particular, if we take some $n<2^k$ we can notice the identity: $$v_2(n)=v_2(2^k-n)=v_2(2n)-1$$ which basically tells us that the function, in the interval $(0,2^{k+1})$ is composed of two copies of itself scaled by half on the $x$-axis and translated $1$ down (one of the copies gets reflected according to the above identity, but the reflection is along a symmetry of the function). Along with the fact that $v_2(2^k)=k$ this defines the function, as well as explains its fractal nature.
Something which you've not noted, but which is also true, is that if we consider the map $$f(x)=\begin{cases}\frac{3x+1}2&&\text{if }x \text{ is odd}\\ \frac{x}2 &&\text{if }x \text{ is even}\end{cases}$$ which is often how the Collatz conjecture is expressed*, then we can get sequences of odd numbers. In fact, noting that $$\frac{3x+1}2=\frac{3}2(x-1)+1$$ we find that iterating that function $n$ times starting from $x$ gives $\left(\frac{3}2\right)^n(x-1)+1$ so the number of times we hit odd numbers is $v_2(x-1)$ under this definition. It should be noted that using both "shortcuts" to define: $$\hat f(x)=\begin{cases}\left(\frac{3}2\right)^{v_2(x-1)}(x-1)+1&&\text{if }x \text{ is odd}\\\frac{x}{2^{v_2(x)}}&&\text{if }x\text{ is even}\end{cases}$$ then it is a result of Simon and De Weger that this map has no cycles of length $2\times 68$ or less (over the positive integers).
I would suggest that the appearance of the 2-adic valuation does not suggest a fractal nature to the problem itself except than to express the fact that the problem relates, somehow to factoring, where such structures naturally arise. Really, what may worry us is the appearance of both $v_2(x)$ and $v_2(x-1)$. This suggests that we care about the factorization both of $x$ and $x-1$ (but the relationship of these factorizations is not well understood). It also suggests a sensitivity to initial conditions - under the first $f$ I gave, we see that if $n$ will be halved many times before hitting an odd number, then $n+1$ will be increased many times before hitting an even number - so the best and worst cases are right next to each other.
(*A prime reason why this is a favored expression is that if we choose some $k$ and some sequence of parities (even/odd), then the portion of positive numbers $n$ such that $n,\,f(n),f^2(n),\,\ldots,\,f^{k-1}(n)$ matches that sequence of parities is asymptotically $\frac{1}{2^n}$. In particular, we can thus imagine that we randomly choose to multiply by $\frac{3}2$ or $\frac{1}2$ at each step - which would tell $f$ should be making things smaller when applied many times.)
I don't know if this answers your question, but it sure is fascinating. The Collatz function has been analytically extended to the complex numbers, and the plot of which points on the complex plane converge and which don't after successive iterations of the extended function (the same technique was famously used on the Mandelbrot set) forms a fractal, and a particularly beautiful one at that. Here are some pictures from the fractal made from the "shortcut" function pointed out by Milo Brandt, which is defined as
$$f(x)=\begin{cases}\frac{3x+1}2&&\text{if }x \text{ is odd}\\ \frac{x}2 &&\text{if }x \text{ is even}\end{cases}$$
extended to the complex numbers as
$$ f(z) = \frac{1}{4} (1 + 4z - (1 + 2z) cos (\pi z)) $$
Centered at the origin with a radius of 3:
Centered at 5+0i with a radius of 5:
An excellent description of how the function was extended can be found here: http://yozh.org/2012/01/12/the_collatz_fractal/ .
As a side note, there's a substantial cluster of convergent points around $e+0i$. I have no idea why.