We define $g:\overline{\mathbb{F}_{p}}\rightarrow\overline{\mathbb{F}_{p}}, x\mapsto x^{d}$ for some $d\in\mathbb{N}$. Let $U = \{z\in\mathbb{F}_{p^{2}}\mid \frac{z+1}{z}\in\mathbb{F}_{p}\} = \mathbb{F}_{p}^{*}\cup U_{p+1}$, where $U_{p+1}$ is the set of elements of $U$ whose order divides $p+1$. Suppose that $z\in U$ such that the order of $z$ is coprime to $d$.
I want to prove that there exists a $m\in\mathbb{N}$ such that $g^{m}(z) = z^{d^{m}} = z$.
Notice that since $z\in U$, we know that $\text{ord}_{z}\mid p-1$ or $\text{ord}_{z}\mid p+1$.
This question comes from studying the periodicity of Chebyshev polynomials on finite fields.
I first read the question as asking for an integer $m$ such that $g^m(z)=z$ holds for all $z\in U$. After a second reading I'm no longer sure. May be it is about a specific element $z$? That changes the situation somewhat.
[The old answer assuming we want the identity to hold for all $z\in U$.]
Such an integer $m$ exists if and only if $\gcd(d,p^2-1)=1$.
If $\gcd(d,p^2-1)$ then the coset of $d$ has finite order $m>0$ in the multiplicative group $\Bbb{Z}_{p^2-1}^*$. So $d^m\equiv1\pmod{p^2-1}$ implying that the desired identity $z^{d^m}=z$ holds for all the elements $z\in \Bbb{F}_{p^2}^*$, and in particular for all the elements of $U$.
On the other hand, if $d$ has a common factor with $p-1$ (resp. with $p+1$), then the restriction of $g$ to $\Bbb{F}_p^*$ (resp. to $U_{p+1}$) is not injective. Therefore the same holds for all the iterations $g^m$.
[Then the update about an individual $z\in U$.]
If we fix an element $z\in U$, then the answer depends on the order of $z$. Let's denote it by $\ell$, so $z^\ell=1$. This time, as observed by the asker, we only require $\gcd(d,\ell)=1$. Anyway, we need the integer $m$ to satisfy the congruence $d^m\equiv1\pmod\ell$. With the assumption on $\gcd$ in place, we know that the residue class of $d$ is an element of the multiplicative group $G=\Bbb{Z}_{\ell}^*$. We can simply use $m=\mathrm{ord}_G(d)$. In other words, the answer exists.