The multiplicative group of integers modulo a prime $p$ is cyclic of order $p-1$. In particular, it contains exactly $\varphi(k)$ elements of order $k$ for each $k$ that divides $p-1$, and contains no elements of order $k$ if $k$ does not divide $p-1$. Here, $\varphi(k)$ is the Euler phi function.
(This because a cyclic group of order $n$ has exactly one subgroup of order $d$ for each divisor $d$ of $n$, and has $\varphi(n)$ generators, i.e., elements of order exactly $n$).
This means the only case in which $\mathbb{Z}_{p}^*$ only has elements of odd order is when $p=2$, when the group is trivial.
The reason your argument does not go through to its conclusion is that if $x$ is of even order $r$, then $x^{r/2}$ is of order $2$. And there is exactly one element of order $2$ in $\mathbb{Z}_p^*$, namely $[-1]$ (the class of $-1$. That means that the factor $x^{r/2}+1$ is actually congruent to $0$ modulo $p$. Also, your conclusion that the factors $x^{r/2}+1$ and $x^{r/2}-1$ “would be two factors of $p$” is incorrect: you have that $p$ divides the product, and hence divides at least one factor. Not that the two factors divide $p$. Indeed, $p$ divides $x^{r/2}+1$ as noted above.
