You are close, as discussed in the comments.
Claim. An abelian group $A$ has a structure as a $\mathbb{Z}[i]$ module if and only if there exists $\phi\in\mathrm{Aut}(A)$ such that $\phi^2(a) = -a$ for all $a\in A$.
Proof. Suppose that $A$ has a $\mathbb{Z}[i]$-module structure. Define $\phi\colon A\to A$ by $\phi(a) = i\cdot a$. Then $\phi(a+b) = i\cdot(a+b) = (i\cdot a)+(i\cdot b) = \phi(a)+\phi(b)$, so $\phi$ is an endomorphism. Since $\phi^4(a) = a$ for all $a\in A$, it follows that $\phi$ is invertible as well, hence an automorphism. Also, $\phi^2(a) = i\cdot(i\cdot a) = (ii)\cdot a = (-1)\cdot a = -a$. Thus, if $A$ has a structure as a $\mathbb{Z}[i]$-module, then there exists $\phi\in\mathrm{Aut}(A)$ such that $\phi^2(a)=-a$.
Conversely, suppose there exists $\phi\in\mathrm{Aut}(A)$ such that $\phi^2(a)=-a$ for all $a\in A$. Define the action by $(r+si)\cdot a = (r+s\phi)(a) = ra+s\phi(a)$. which is easily verified to be a module structure. QED
From this it easily follows that if $-1$ is a square in $\mathbb{F}_p$, then we can make $\mathbb{F}_p$ into a $\mathbb{Z}[i]$ module by considering the automorphism of $\mathbb{F}_p$ (as an abelian group) given by multiplication by $\alpha$, where $\alpha^2 = -1$. It also shows, via your argument, that $\mathbb{F}_p^2$ is a $\mathbb{Z}[i]$ module, by considering $\phi(a,b) = (-b,a)$.
You still have to show the necessity for $\mathbb{F}_p$, though.
Hint: See what $i\cdot 1$ is.