I have read about a theorem of Siegel on this Terence Tao's blogpost :
Theorem (Siegel’s theorem on integer points) Let ${P \in {\bf Q}[x,y]}$ be an irreducible polynomial of two variables, such that the affine plane curve ${C := \{ (x,y): P(x,y)=0\}}$ either has genus at least one, or has at least three points on the line at infinity, or both. Then ${C}$ has only finitely many integer points ${(x,y) \in {\bf Z}^2}$.
Question : conversely, if the genus is $g\leq0$ does that imply that the number of integer points is necessarily infinite ?
For example, consider the curve $y^2=ax^2+bx+c$ with $a,b,c\in\mathbb{N}$ and the discriminant of the RHS polynomial in $x$ is $\Delta>0$.
If I understand well, it has degree $2$, so it has genus $g=\frac{(2-1)(2-2)}{2}-s=-s$, where $s$ is the "number of singularities, properly counted".
So here $g\leq 0$. Is the number of integer points then infinite for all those curves ?
No, if $g=0$, then the number of points with integer coordinates is not necessarily infinite. For example, $x^2+y^2=1$ only has four points with integer coordinates, namely $(\pm 1,0)$ and $(0,\pm 1)$. However, other curves of genus $0$ have infinitely many integral points, for example $x^2-2y^2=1$ has infinitely many points with integer coordinates.
In the particular family that you are considering, one can take for example $y^2=7x^2+14x+3$ (with discriminant $14^2-4\cdot 21>0$), and show there are no integral points in the family. (If there was an integral point $(x_0,y_0)$, then reducing modulo $7$ we would reach a contradiction, because $y_0^2\equiv 3 \bmod 7$ is impossible.)