What is information about the existence of rational points on hyperelliptic curves over finite fields available?

[Edited to remove material subsumed and improved by Felipe's answer.] Here is some historical info. Dickson studied this question in his 1909 paper "Definite forms in a finite field". For Dickson, a "definite form" is a homogeneous $f(x,z)\in\mathbb{F}_q[x,z]$ which takes nonzero square values for all $(x,z)$ in $\mathbb{F}_q\times\mathbb{F}_q$ except $(0,0)$. If $q$ is odd and $f(x,z)$ is not a square then Dickson's condition is equivalent to saying that the hyperelliptic curve $y^2=f(x,1)$ has $2q+2$ points over $\mathbb{F}_q$, or equivalently, its quadratic twist $y^2=nf(x,1)$ has no points (where $n$ is any nonsquare in $\mathbb{F}_q$). In modern language, Dickson showed that there are no pointless genus$2$ curves over $\mathbb{F}_q$ if $q$ is odd and $q\ge 13$. Carlitz took up this topic in a series of papers, and among other things made the connection with Weil's bound, which implies that a pointless hyperelliptic curve over $\mathbb{F}_q$ has genus at least $(q+1)/(2\sqrt{q})$, or roughly $\sqrt{q}/2$. As Felipe's answer indicates, this bound is essentially best possible when $q$ is an odd square. It can be improved by a factor of roughly $\sqrt{2}$ (and possibly much more) when $q$ is prime. It is known that genus$2$ pointless hyperelliptic curves exist over $\mathbb{F}_q$ precisely when $q\le 11$, and in genus$3$ the analogous result is $q\le 25$ (the latter is due to Howe, Lauter, and Top). Further experimental results over small prime fields appear in papers by Glazunov. 


As Mike says, there isn't much beyond the Weil bound. For prime fields, there is a slight improvement due to Stark. If the field has $q=r^2$ elements with $q$ odd, then one can find $a,b$ such that $y^2=ax^{r+1}+b$ has no points, which shows that you cannot improve the Weil bound. For arbitrary $q$ odd, there is this beautiful idea of I. Shparlinski: For a monic irreducible polynomial $f$ of degree $d$, the vector of values of $f$ modulo squares is one of $2^q$ possibilities. There are about $q^d/d$ choices for $f$ and so if $q^d/d > 2^q$ or thereabouts, you get two monic irreducibles $f,g$ such that $fg$ takes only square values and so, if $c$ is a nonsquare, $y^2=cf(x)g(x)$ has no points and you can take $d$ about $q/\log q$. I don't know how to do better than this. 

