smc

Plug in $y=mx+1$ into the ellipse's equation to find the intersection points: $$x^2+4(mx+1{)}^2=1$$ After simplifying, we have a quadratic in $x$: $$(4m^2+1)x^2+8mx+3=0$$ Because there is only one intersection point, then the quadratic has only one solution. This can only happen when the discriminant is 0. $$\Delta =b^2-4ac=(8m{)}^2-(4)(1+4m^2)(3)=0$$ Solving, we find $m^2=\boxed{\frac{3}{4}}.$