I'm trying to prove that a finite field of size $q$ cannot be algebraically closed. I know I can prove it by showing that there are no roots to the polynomial $x^q-x+1$, but what about the polynomial $x^{q+1}+1$? Is there any relationship between the two polynomials?
Since for a finite field $F$, we have that $x^q=x$, so does $x^{q+1}+1=x^2+1$?
It's not true that there are no roots of $x^{q+1}+1$, at least not in general. As you say, since $a^q=a$ for all $a\in \mathbb{F}_q$, this is equivalent to asking for roots of $x^2+1$ (at least as long as you restrict to roots that are in $\mathbb{F}_q$). A root of $x^2+1$ is just a square root of $-1$, or a primitive $4$th root of unity (if the characteristic is not $2$). So such a root exists whenever $q-1$ is divisible by $4$. (Or, such a root exists in any field of characteristic $2$, since $x=1$ is a root!)