This question asks us to find the number of solutions to a system with one linear equation and one quadratic equation. Since we have a line ($y = 2x + 3$) and a parabola ($y = x^2 + 1$), we solve by setting the equations equal: $2x + 3 = x^2 + 1$. Rearranging gives $x^2 - 2x - 2 = 0$, and using the quadratic formula yields $x = \frac{2 ± \sqrt{4 + 8}}{2} = \frac{2 ± \sqrt{12}}{2} = 1 ± \sqrt{3}$. Since we get two distinct real solutions for $x$, there are exactly 2 intersection points. Remember that a line and parabola can intersect at 0, 1, or 2 points depending on whether the discriminant is negative, zero, or positive.