Show if $2$ and $1+\sqrt{-5}$ belong to the same principal ideal $I$ of $\mathbb{Z}[\sqrt{-5}]$ then $I=\mathbb{Z}[\sqrt{-5}]$. 
Show if $2$ and $1+\sqrt{-5}$ belong to the same principal ideal $I$ of $\mathbb{Z}[\sqrt{-5}]$ then $I=\mathbb{Z}[\sqrt{-5}]$.

I have proved so far that 2 and $1+\sqrt{-5}$ is irreducible and they are not associates.
 A: This answer uses the information, given in OP, that $2, 1 + \sqrt{-5}$ are irreducible elements that are not associated. 
Suppose $I = (a)$. If $\{2, 1 + \sqrt{-5}\} \subset I $ then $a \mid 2$ and $a \mid 1 + \sqrt{-5}$. Thus as $2, 1 + \sqrt{-5}$ are irreducible:


*

*either $a$ is a unit, 

*or $a$ is associated to $2$ and $a$ is associated to $1 + \sqrt{-5}$.


Yet the second part would imply that $2$ and $1 + \sqrt{-5}$ are associated, which is not the case. So $a$ is a unit and you are done.

Note that this argument works for any two irreducible and non-associated elements in a domain. If a principle ideal $(a)$ contains two elements $b_1,b_2$ then $a \mid b_1$ and $a \mid b_2$. So, $a$ is a common divisor of $b_1$ and $b_2$. Yet, if $b_i$ is irreducible, then the only divisors are units and associates. Thus, if $b_1$ and $b_2$ are not assocaites, only units are left as common diviors.
A: Assume that $2$ and $1+\sqrt{-5}$ are in an ideal generated by $a+b\sqrt{-5}$.
Let's start with the equation
$$2=(a+b\sqrt{-5})(c+d\sqrt{-5})$$
take modules and square:
$$4=(a^2+5b^2)(c^2+5d^2)$$
Now it is clear that $b=d=0$, and $a^2c^2=4$.
But, on the other hand,
$$1+\sqrt{-5}=(a+b\sqrt{-5})(c'+d'\sqrt{-5})=ac'+ad'\sqrt{-5}$$
This yields:
$$ac'=1$$
$$ad'=1$$
So $|a|=1$. and therefore, the generator of our ideal is $1$ or $-1$.
A: I just thought that I would recast ajotatxe’s argument in a slightly different manner.


*

*Observe that we have a multiplicative function $ N: \Bbb{Z}[\sqrt{-5}] \to \Bbb{N}_{0} $ defined by
$$
\forall (a,b) \in \Bbb{Z}^{2}: \quad
N(a + b \sqrt{-5}) \stackrel{\text{df}}{=} a^{2} + 5 b^{2}.
$$

*Suppose that $ \{ 2,1 + \sqrt{-5} \} \subseteq \langle x \rangle $ for some $ x \in \Bbb{Z}[\sqrt{-5}] $.

*As $ N(2) = 4 $, we have $ N(x) | 4 $, and as $ N(1 + \sqrt{-5}) = 6 $, we have $ N(x) | 6 $.

*Hence, $ N(x) \in \{ 1,2 \} $.

*It is easily verified, however, that $ 2 $ is not in the range of $ N $, so we must have $ N(x) = 1 $.

*Therefore, $ x = \pm 1 $, which yields $ \langle x \rangle = \mathbb{Z}[\sqrt{-5}] $.

