So the only common divisor of \(\displaystyle 1 + \sqrt{-5}\) and 2, in the ring \(\displaystyle Z\sqrt{-5}\) is 1?

No. Any common divisor will have norm 1, and \(\displaystyle N(a) = 1 \Leftrightarrow a\) is a unit. This may be in your notes (the \(\displaystyle N(a) = 1 \Leftrightarrow a\) is a unit bit). Otherwise, it is good exercise to prove it.

You need to prove that if a|2 and a| \(\displaystyle (1 + \sqrt{-5})\) then N(a)=1. To do this, you must notice that as \(\displaystyle N(2) = N(1 + \sqrt{-5})=4\) and so if a|2 and a|\(\displaystyle (1 + \sqrt{-5})\) then N(a)|4. There are only three numbers which divide 4. They are 1, 2 and 4.

If N(a)=1 then \(\displaystyle a\) is a unit.

N(a)=2 cannot happen, you must prove this.

If N(a)=4 then you need to prove that \(\displaystyle a\) cannot divide both ring elements. To do this, notice that there must exist \(\displaystyle b_1, b_2\) such that \(\displaystyle ab_1 = 1 + \sqrt{-5}\) and \(\displaystyle ab_2 = 2\) \(\displaystyle \Rightarrow N(ab_1) = N(ab_2) = 4 \Rightarrow N(b_1) = N(b_2) = 1 \Rightarrow 2 {b_{2}}^{-1}b_1 = 1 + \sqrt{-5}\). Is this possible?