Question
A weak form of Sharkovskii's theorem in $1$D dynamical systems states that, if a continuous function $f:I\to I$ does not include a periodic point of least period $2$ on $I$, then there is no periodic point of least period $3$ on $I$. (This can be strengthened to the conclusion that every periodic point on $I$ is a fixed point.)
Are there generalizations of this result to dynamical systems in $\mathbb{R}^2$?
Motivation
One of my first answers addressed a seemingly-simple algebra problem: Given a particular function $f(x)$, are there distinct real $x,y,z$ which solve the system of equations
$$y=f(x),\quad z=f(y),\quad x=f(z)\quad?$$
This amounts to finding real $x$ such that $x=f^{3}(x)$ but $x\neq f(x)\neq f^2(x)$ (in the sense of function composition); in other words, $x$ is a periodic point of least period $3$. The advantage of this formulation is that we can apply Sharkovskii's theorem to deduce the following criterion: If $f(x)$ has no real periodic point of least period $2$, then neither does it have a real periodic point of least period $3$.
Thus a necessary condition for non-trivial solutions of the system of equations above is that a non-trivial solution exist for the system $x=f(y),\quad y=f(x)$. In fact, Sharkovskii's theorem furthermore implies that the same is true for every such multiplet of equations. This means that, by ruling out real primitive $2$-cycles, we have the possibility to immediately rule out the solution of all such systems of equations. That is rather powerful...
What I wonder is whether this approach is applicable to a variant of such systems. Suppose we instead consider a continuous map $f:\mathbb{R}^2\to \mathbb{R}^2$ and the system of equations $$z=f(x,y),\quad x=f(y,z),\quad y=f(z,x).$$ We may express this in terms of periodic points via the map $F(x,y)=(f(x,y),y)$, for then the above system is equivalent to $$(z,y)=F(x,y),\quad (y,x)=F(z,x),\quad (x,z)=F(y,z).$$ Thus a solution with distinct real $x,y,z$ would be a periodic point of least period $f$ for the $2$D mapping $f(x,y)$.
Alas, Sharkovskii's theorem applies only to continuous $1$D mappings. So, unlike the first case, there is no obvious necessary condition for the existence of a $3$-cycle. To overcome this would require some version of Sharkovskii's theorem to hold in two dimensions. So some bridging of my ignorance would be appreciated.