I can show that $I$ itself is not a principal ideal by using the multiplicative norm $N(a+b\sqrt{-10}) = a^2 + 10 b^2$ using the following:
Suppose that $I=(\alpha )$. Then $N(\alpha) | 2$ and $N(\alpha)|10$, so that $N(\alpha ) | \gcd(2,10) =2$. Thus $\alpha$ must be $1$, or $2$. Then write $\alpha = a + b\sqrt{-10}$, and we get $a^2 + 10 b^2 = 1$ or $=2$. The only possible solution for $a,b \in \mathbb{Z}$ is $\alpha = \pm 1$. This however cannot be right as $\pm 1 \in I$. So the ring is not a principal ideal domain
Now, I work out what $I^2$ is. I get $$ I^2 = \left\{ \sum (2a + b \sqrt{-10})(2c + d\sqrt{-10}) \right\} = \left\{ \sum 4ac - 10bd + 2(cb+ad)\sqrt{-10} \right\}.$$ Now the question is, does this ideal have a single generator?
The ideal $I^2$ is the product of $I$ with itself. Hence it is generated by the products of pairs of generators of $I$, meaning that $$I^2=(2\times2,2\times\sqrt{-10},\sqrt{-10}\times\sqrt{-10}).$$ This can be simplified to $I^2=(4,2\sqrt{-10},-10)=(2,2\sqrt{-10})=(2)$.