jmc

If $m>0,$ then the graph of $x^2+m{y}^2=4$ is an ellipse centered at the origin. The endpoints of the horizontal axis are $(\pm 2,0),$ while the endpoints of the vertical axis are $\left(0,\pm \frac{2}{\sqrt{m}}\right).$ If $m<1,$ then the vertical axis is longer, so it is the major axis, and the distance from the foci to the origin is $$\sqrt{{\left(\frac{2}{\sqrt{m}}\right)}^2-2^2}=\sqrt{\frac{4}{m}-4}.$$Since the foci lie on the circle $x^2+y^2=16,$ which has radius $4$ and is centered at the origin, we must have $$\sqrt{\frac{4}{m}-4}=4$$which gives $m=\frac{1}{5}.$ If $m>1,$ then the horizontal axis is longer, so it is the major axis. But the endpoints of the horizontal axis are $(\pm 2,0),$ so it is impossible that the foci of the ellipse are $4$ units away from the origin in this case.

If $m<0,$ then the graph of $x^2+m{y}^2=4$ is a hyperbola centered at the origin, with the vertices on the $x-$axis. Its standard form is $$\frac{x^2}{2^2}-\frac{y^2}{{\left(\sqrt{-\frac{4}{m}}\right)}^2}=1,$$so the distance from the foci to the origin is $$\sqrt{2^2+{\left(\sqrt{-\frac{4}{m}}\right)}^2}=\sqrt{4-\frac{4}{m}}.$$Therefore, we must have $\sqrt{4-\frac{4}{m}}=4,$ which gives $m=-\frac{1}{3}.$

Therefore, the possible values of $m$ are $m=\boxed{\frac{1}{5},-\frac{1}{3}}.$