Why does the difference equation $x_nx_{n+2}=w^5x_n-(w^2+w^3)x_{n+1}+x_{n+2}$ generate cyclic sequences?

The substitution $z_n=x_n−1$ really helps us, it leads to the equation $$z_nz_{n+2}=az_{n+1}+b,$$ where $a=-\omega^2-\omega^3, b=1-\omega^2-\omega^3.$ We can easily see that if $z_n=u, z_{n+1}=v$ then $$z_{n-1}=\frac{au+b}{v}, z_{n+2}=\frac{av+b}{u},$$ and $$z_{n-2}=\frac{a^2u+bv+ab}{uv}, z_{n+3}=\frac{a^2v+bu+ab}{uv},$$ so if $a^2=b$ then $z_n$ is $5$-periodic, so and $x_n$. It is true because $$a^2=(\omega^2+\omega^3)^2=2+\omega+\omega^4=1+(1+\omega+\omega^4)=1-\omega^2-\omega^3=b.$$

Too long for a comment: This is to show one periodic sequence that satisfies a similar recurrence in a natural way.

The recurrence $x_nx_{n+2}=x_n+tx_{n+1}+x_{n+2}-(1+t)$ is 6-periodic, for every $t \neq 0$ (and non-degenerate initial values).

We start by observing that the sequence $u,v,v/u,1/u,1/v,u/v,u,v,\ldots$ is 6-periodic, and satisfies the recurrence $y_ny_{n+2}=y_{n+1}$. Now, we can check that the recurrence $y_ny_{n+2}=ry_{n+1}$ is also periodic for any non-zero choice of $r$. In this latter recurrence, substitute $y_n=r(x_n-1)$ and $r=1/t$ to get the first recurrence stated above.

The OP's first recurrence also simplifies in form with the substitution $y_n=x_n-1$, but I don't see a similar natural construction for period 5 (or other periods).