If $ m > 0, $ then the graph of $ x^2+my^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+my^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}} $.