I've been able to show this for the first few:
When $p=2$, $f=x^2+x+1$ does not have a root in $\mathbb{Z}_2$
When $p=3$, $f=x^2-2$ does not have a root in $\mathbb{Z}_3$
When $p=5$, $f=x^2-3$ does not have a root in $\mathbb{Z}_5$
When $p=7$, $f=x^2-5$ does not have a root in $\mathbb{Z}_7$
It looks like it follows a recursion, but I'm not sure how to go about showing this for all $p$. Any hints?
In fact, a stronger statement is true: No finite field is algebraically closed. Let $\{a_i\}$ be all of the elements in your finite field, and consider $1+\prod_1^n(x-a_i)$. This polynomial does not have a root in the field, so the field is not algebraically closed.
Assume $p\ge 3$. By Fermat's little theorem then we get $x^{p-1}=1$ for all $x\neq 0$. Then what is the solution of $x^{p-1}=2$?
For $p$ odd, there are $\frac{p-1}{2}$ quadratic non-residues, so there are quadratic polynomials that have no root in $\mathbb{Z}_p$.
This argument is similar to Andre's:
Let
$$A := \{ x^2+ax+b | a,b \in \mathbb Z_p \}$$ $$B:= \{ (x-a)(x-b) | a,b \in \mathbb Z_p \}$$
They are bot finite sets.
The function $F : A \to B$ defined by $F(x^2+ax+b)=(x-a)(x-b)$ is onto but not one to one. Thus, $B$ has strictly less elements than $A$.
Thus, $B \subset A, B \neq A$. Hence we can find some Polynomial $P \in A \setminus B$. But such a polynomial cannot have roots in $\mathbb Z_p$.
