jmc

Let $z=x+yi,$ where $x$ and $y$ are real. Then $$(x+yi{)}^4=x^4+4i{x}^{3}y-6x^2{y}^2-4ix{y}^3+y^4=-4.$$Equating real and imaginary parts, we get $$\begin{array}{rl}{x}^4-6x^2{y}^2+y^4& =-4,\\ 4x^{3}y-4x{y}^3& =0.\end{array}$$From the equation $4x^{3}y-4x{y}^3=0,$ $4xy(x^2-y^2)=0.$ If $x=0,$ then $y^4=-4,$ which has no solutions. If $y=0,$ then $x^4=-4,$ which has no solutions. Otherwise, $x^2=y^2.$

Then the first equation becomes $-4x^4=-4,$ so $x^4=1.$ Hence, $x=1$ or $x=-1.$ In either case, $x^2=1,$ so $y^2=1,$ and $y=\pm 1.$ Therefore, the solutions are $\boxed{1+i,1-i,-1+i,-1-i}.$