The point $(-2m,m+1)$ is an interior point of the smaller region bounded by circle $x^2+y^2=4$ and the parabola $y^2=4x$ then:

The point $(–2k, k + 1)$ is the interior point of the circle and parabola

$So {\left(2k\right)}^2+{(k+1)}^2-4<0\Rightarrow 4k^2+k^2+2k+1-4<0\Rightarrow 5k^2+2k-3<0$

$(k+1)(k-\frac{3}{5})<0\Rightarrow k\in (-1,\frac{3}{5})\to \left(1\right)$

$now,{(k+1)}^2-4(-2k)<0$

$k\in (-5-2\sqrt{6},-5+2\sqrt{6})\to \left(2\right)$

$Sofrom\left(1\right)\&\left(2\right)k\in (-1,-5+2\sqrt{6})$