I thought about this years ago but never had it resolved. Sharkovskii's Theorem has a corollary according to Wikipedia that if there exists a 3-cycle on a map on $\mathbb{R}$, then the map has $n$-cycles for every $n\in\mathbb{N}$. Now, the Collatz map $f(x)$ has a 3-cycle: $f(f(f(4))) = 4$...
Now, I assume the conclusion that the Collatz map has cycles of every size is wrong (otherwise, it would have been solved) but why does it not apply here?
In short, the Collatz conjecture does have cycles of every period, including $3$, but this is over the 2-adic numbers. It is, by and large, topologically conjugate to the Dyadic Transformation.
Two functions are topologically conjugate to each other when there is a homeomorphism which conjugates one to the other. A consequence of topological conjugacy is that functions which are topologically conjugate share the same orbits.
A little more precisely - the Dyadic map has transitive orbits of period $n$ which are in one to one bijection with the base $2$ Lyndon words of length $n$. If one converts the Dyadic Transformation to the Cantor set by writing every binary number in the unit interval with twos in place of ones, and reinterpret it as a base $3$ number, the Dyadic map remains pretty much the same except it now has one additional orbit terminating in a fixed point. This function is now exactly topologically conjugate to the Bit Shift map on 2-adic numbers, which is given by $x\mapsto (x-1)/2$ for odd $x$ and $x\mapsto x/2$ for even $x$.
The bit shift map is topologically conjugate to any other shift map on the 2-adic numbers where you replace the $(x-1)$ part with $ax-b$, where $a,b$ are two 2-adic units (i.e. odd numbers). They all have orbits of the full gamut of periods, enumerated by the length $n$ base $2$ Lyndon words.
The challenge is to show that the unique period $2$ orbit is the only one which the natural numbers fall into.
EDIT
Sorry this is an edit following the downvote, in case this much wasn't clear from the above... the function $x\mapsto x+2^{\nu_2(x)}$ shares precisely the same transitive orbits as the bit shift map, and is continuous on the 2-adic numbers. Then by the topological conjugacy given above you get a homeomorphism to the Cantor set, then using the devil's staircase you have a homeomomorphism from the Collatz conjecture to a continuous endomorphism on a segment of $\Bbb R$, hence the system having periodic points of period $3$ and every period by Sharkovskii's theorem.