Question: So far I have it down to

\begin{align}
\sqrt{x+iy}&=u+iv \\
\Rightarrow x+iy&=u^2-v^2+i(2uv) \\
\Rightarrow x&=u^2-v^2, y=2uv 
\end{align}
But I keep getting stuck here.

Answer: 
You have arrived at $x = u^2-v^2$ and $y = 2uv$. Now use the second relation to eliminate either $u$ or $v$ from the first relation, and you obtain expressions for $u$ or $v$ in terms of $x$ and $y$.

From $4u^4-4xu^2-y^2=0$ we get: $u^2 = 0.5(x+\sqrt{x^2+y^2})$

From $4v^4+4xv^2-y^2=0$ we get: $v^2 = 0.5(-x+\sqrt{x^2+y^2})$

If you take the difference of these two results or multiply them you get the original equations back. If you take the sum, you see that the $(u,v)$ norm is simply the square root of the $(x,y)$ norm.

The final step to obtain $u$ and $v$ is to take the square root and choose the correct sign, so that their product satisfies $2uv = y$. The standard choice is to take $u$ greater or equal to zero, and to give $v$ the same sign as $y$. But minus this result is equally valid.
You are almost done, following your way we obtain, with $u\neq 0$

$$2uv=y \implies v = \frac y {2u}$$

and then from the first equation

$$x=u^2-\left(\frac y {2u}\right)^2 \implies 4u^4-4u^2x-y^2=0 \implies u^2=\frac{x\pm\sqrt{x^2+y^2}}{2}$$

and then by $r=\sqrt{x^2+y^2}$

$$u=\pm \sqrt\frac{x+r}{2},\; v= \pm \frac y 2 \sqrt\frac 2 {x+r}$$

The case $u=0$ can be checked by inspection.