How many points does the surface $$\mathbb{H} = \{(z_1,z_2,z_3,z_4,z_5)\in \mathbb{F}_{p^2}: z_1 - z_1^p - z_2z_3^p + z_2^pz_3 - z_4z_5^p + z_4^pz_5 = 0\}$$ contain in $\mathbb{F}^5_{p^2}$? Furthermore, is it possible to determine how many lines does that surface contain?
What might be of a help is that if we have $\Phi(a) = a^p$, then for $\mathbb{F}_{p^2}$ the map $\Phi$ is an involution, e.g. $\Phi^2(a) = a$.
Counting the number of points is simple. I need the trace function $$ tr:\Bbb{F}_{p^n}\to\Bbb{F}_p, x\mapsto x+x^p+x^{p^2}+\cdots+x^{p^{n-1}} $$ and its properties $$ \begin{aligned} tr(x+y)&=tr(x)+tr(y)\\ tr(x^p)&=tr(x) \end{aligned} $$ that hold for all $n$ and all $x,y\in\Bbb{F}_{p^n}.$
Point count follows from the following standard fact.
Proposition. The equation $$ z^p-z=u $$ has $p$ solutions $z\in\Bbb{F}_{p^n}$ if $tr(u)=0$, and no solutions otherwise. Furthermore, if $z=\xi$ is one solutions, all the solutions are of the form $z=\xi+a, a\in\Bbb{F}_p$.
Let us rewrite your equation in the form $$z_1^p - z_1 =- z_2z_3^p + z_2^pz_3 - z_4z_5^p + z_4^pz_5.\qquad(*)$$ Consider the right hand side. For any choices of $z_2,z_3\in \Bbb{F}_{p^2}$ we have $$ \begin{aligned} tr(-z_2z_3^p+z_2^pz_3)&=-tr(z_2z_3^p)+tr(z_2^pz_3)\\ &=-tr(z_2z_3^p)+tr(z_2^{p^2}z_3^p)\\ &=-tr(z_2z_3^p)+tr(z_2z_3^p)=0, \end{aligned} $$ because, as you pointed out, $z_2^{p^2}=z_2$.
Similarly $$ tr(- z_4z_5^p + z_4^pz_5)=0 $$ for all $z_4,z_5\in \Bbb{F}_{p^2}$. So for all values of $z_2,z_3,z_4,z_5$ ($p^8$ choices) the trace of the right hand side of $(*)$ vanishes. Therefore the Proposition implies that the equation has exactly $p^9$ solutions $(z_1,z_2,z_3,z_4,z_5)\in\Bbb{F}_{p^2}^5.$