jmc

We attempt to solve the system $y=x^2+2$ and $y^2-m{x}^2=1.$ The first equation gives $x^2=y-2,$ so we can substitute into the second equation to get $$y^2-m(y-2)=1,$$or $$y^2-my+(2m-1)=0.$$For the parabola and hyperbola to be tangent, this equation must have exactly one solution for $y,$ so the discriminant must be zero: $$m^2-4(2m-1)=0.$$Thus, $m^2-8m+4=0,$ which gives $$m=\frac{8\pm \sqrt{8^2-4\cdot 4}}{2}=4\pm 2\sqrt{3}.$$To choose between the two possible values of $m,$ we attempt to solve for $y$ in the equation $y^2-my+(2m-1)=0.$ For $m=4\pm 2\sqrt{3},$ we have $$y=\frac{m\pm \sqrt{m^2-4(2m-1)}}{2}=\frac{m}{2},$$because these values of $m$ make the discriminant zero. Since $y=x^2+2,$ we have $y\ge 2,$ so we must have $\frac{m}{2}\ge 2,$ or $m\ge 4.$ Therefore, we must choose the root $m=\boxed{4+2\sqrt{3}}.$ (Note that only the top branch of the hyperbola is shown below, in blue.)