smc

Substituting $y=A{x}^2$ into the equation $y^2+3=x^2+4y$ gives $$\begin{array}{rl}\left(A{x}^2\right)+3=x^2+4\left(A{x}^2\right)& ⟺A^2{x}^4+3=x^2+4A{x}^2\\ & ⟺A^2{x}^4-\left(4A+1\right)x^2+3=0\\ & ⟺x^2=\frac{4A+1\pm \sqrt{4A^2+8A+1}}{2A^2}\\ & \text{(using the quadratic formula)}.\end{array}$$ Now observe that since $A$ is positive, $4A^2+8A+1$ is also positive, so the square root will always give two distinct real values. Moreover, $${\left(4A+1\right)}^2=16A^2+8A+1>4A^2+8A+1,$$ so $4A+1-\sqrt{4A^2+8A+1}>0$, meaning that both solutions for $x^2$ are positive. Hence both solutions will give $2$ distinct values of $x$ (the positive and negative square roots), and each of these will correspond to a distinct point of intersection of the graphs, so there are $2\cdot 2=\boxed{\text{exactly }4}$ points of intersection.