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,  ${(k+1)}^2-4(-2k)<0$

$\Rightarrow k^2+2k+1+8k<0\Rightarrow k^2+10k+1<0$

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

So from (1) & (2) $k\in (-1,-5+2\sqrt{6})$