jmc

Note that $z^2+4=(z+2i)(z-2i),$ so we can write the given equation as $$|z+2i||z-2i|=|z||z+2i|.$$If $|z+2i|=0,$ then $z=-2i,$ in which case $|z+i|=|-i|=1.$ Otherwise, $|z+2i|\ne 0,$ so we can divide both sides by $|z+2i|,$ to get $$|z-2i|=|z|.$$This condition states that $z$ is equidistant from the origin and $2i$ in the complex plane. Thus, $z$ must lie on the perpendicular bisector of these complex numbers, which is the set of complex numbers where the imaginary part is 1.



In other words, $z=x+i$ for some real number $x.$ Then $$|z+i|=|x+2i|=\sqrt{x^2+4}\ge 2.$$Therefore, the smallest possible value of $|z+i|$ is $\boxed{1},$ which occurs for $z=-2i.$