jmc

We can factor $x^4+64$ as a difference of squares: $$\begin{array}{rl}{x}^4+64& =(x^2{)}^2-(8i{)}^2\\ & =(x^2-8i)(x^2+8i).\end{array}$$Thus, the solutions are the square roots of $8i$ and $-8i$.

The square roots of $i$ are $\pm \left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\right)$. Thus, the square roots of $8i$ are $\pm \sqrt{8}\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\right)=\pm (2+2i)$, and the square roots of $-8i$ are $\pm \sqrt{-8}\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\right)=\pm (2i-2)$.

Therefore, the solutions of the original equation are $x=\boxed{2+2i,-2-2i,-2+2i,2-2i}$.