How do you square an ideal?

Both other answers are right but fail to give the answer in its simplest form. Note that $$-4+(2 + 2 \sqrt{-5}) - (-4+2 \sqrt{-5}) = 2.$$ So $2$ is in the ideal $\langle 4, (2 + 2 \sqrt{-5}), (-4+2 \sqrt{-5}) \rangle$ and, conversely, it is easy to see that each of $4$, $2 + 2 \sqrt{-5}$ and $-4+2 \sqrt{-5}$ is divisible by $2$. So $$\langle 2, 1 + \sqrt{-5} \rangle^2 = \langle 4, (2 + 2 \sqrt{-5}), (-4+2 \sqrt{-5}) \rangle = \langle 2 \rangle.$$

When you square an ideal you're really looking at products of two things coming from your ideal. In you case $\langle 2, 1+\sqrt{-5}\rangle^2$ means take something in $\langle 2, 1+\sqrt{-5}\rangle$ and multiply it by something else in the ideal; the set of all the things you get by doing this is the ideal $\langle 2, 1+\sqrt{-5}\rangle^2$.

In order to work with this you can just multiply your generators together, so $\langle 2, 1+\sqrt{-5}\rangle^2=\langle 4,2+2\sqrt{-5},-4+2\sqrt{-5}\rangle$. Then you might look for relations between these generators to reduce this set. In this case it does not appear that there are any relations between them.

You are correct that in a P.I.D., $\langle a \rangle \langle b\rangle=\langle ab\rangle$ provided $a,b$ aren't units (in which case $\langle a \rangle$ or $\langle b \rangle$ would be the entire ring).

For a P. I. D., yes. For the general case, the product of two ideals $I=(a_1,\dots, a_n)$ and $J=(b_1,\dots, b_p)$ is the ideal generated by all possible products $a_ib_j$.

Hence the square of $I=(2, 1+\sqrt{-5})$ is the ideal $I^2=(4,2+2\sqrt{-5},-4+2\sqrt{-5})$.