Pointless, non-singular, absolutely irreducible affine plane curves over finite fields We think the following is true:
For all sufficiently large primes $p$ and all natural $g \ge 1$, there
exists affine plane curve $f(x,y)=0$ over $\mathbb{F}_p$ which 
is non-singular, absolutely irreducible, of genus $g$ and it doesn't have
any rational points.
Is it true? 
Is it known?

This doesn't violate Hasse-Weil bound, because the bound
requires SMOOTH projective model and our examples have very
few singular points on the projective model.
For $p=13,g=1$ check this question and comments
Added Example of pointless non-sigular affine curves for $g=1$ defined
by two equations.
Let $f_0=x^3+y^2-1,f_1=z(x^p-x)-1$ and the curve $C : f_0=0,f_1=0$.
Then $f_1$ is linear in $z$ so $C$ is birationally equivalent to $f_0=0$.
The curve is pointless because $x^p-x$ is zero modulo $p$.
To get a single equation for the curve set $f$ the resultant of $f_0$ and $f_1$ wrt
$x$, experimentally it is irreducible.
 A: Yes, it is true.
If you have any smooth geometrically integral plane affine curve $C/\mathbb{F}_q$ of genus $g$ with $|C(\mathbb{F}_q)|<q^2$, then you can construct such a curve $C'$ birational to $C$ with $|C'(\mathbb{F}_q)|=0$:
Remove the rational points one by one by pushing them out to infinity: Pick $P\in C(\mathbb{F}_q)$ and $Q\in\mathbb{A}^2(\mathbb{F}_q)\setminus C$. By a linear transformation assume that $Q=(0,0)$ and $P$ lies on the line $L=\{X=0\}$. Now let $C_0=C\setminus L$ and observe that $(x,y)\mapsto(x,x^{-1}y)$ is an isomorphism of $C_0$ onto another plane affine curve. This curve is birational to the original curve and is still smooth but has at least one rational point less. Now repeat.
That a smooth geometrically integral plane affine curve $C/\mathbb{F}_q$ of genus $g$ with $|C(\mathbb{F}_q)|<q^2$ (i.e. that is not space filling) exists is clear.
(Remark: I agree that the statement above sounds a bit odd, as we are pushing up to ${\rm min}(q^2-1,q+1+2g\sqrt{q})$ many rational point to the line at infinity, which has only $q+1$ many rational points. But note that the points at infinity can be highly singular.)
A: Partial results, because of contradicting comments in the question.
Let $f(x,y)=0$ be smooth and absolutely reducible curve over $\mathbb{F}_p$.
To reduce the number of points while keeping the genus the same,
for $f_1, f_2 \in \mathbb{F}_p[x,y]$ and a new variable $z$
set $F(x,y,z)=f_1 z - f_2$ and take the curve $f=0,F=0$. 
This is linear in $z$, so it preserves the genus of $f=0$. 
If you want only one equation for the curve, take $G$ to be
the resultant of $f$ and $F$ with respect to $x$.
We have the constraints $f_1(x,y) = 0 \implies f_2(x,y)=0$, which may fail
by construction.
Extreme example $p=13,f=x^3+y^3-2,f_1 = x^p-x, f_2 =1$, the plane curve $G=0$ is pointless
and smooth.
since $x^p-x=0$ for all $x$.
where
G=y^39*z^3 - 3*y^27*z^3 + 2*y^24*z^3 - 3*y^21*z^3 + y^18*z^3 + 2*y^12*z^3 - 5*y^9*z^3 - y^6*z^3 - 4*y^3*z^3 - 3*z^3 + 1
We think this $G$ is absolutely irreducible over $\mathbb{F}_{13}$.
If we want to exclude a set $S$ of $x$ coordinates, take 
$f_2=1, f_1=z \prod_{a \in S}(x-a)$.
Other $f,f_1,f_2$ may give non singular plane curves with few or no points,
while keeping $G$ non-singular.
