Prove that $p^2$ is the principal ideal $(2)$.

Here’s another approach:
To multiply two ideals, say $I=(a_1,a_2,\cdots,a_m)$ and $J=(b_1,\cdots,b_n)$, where what’s in each pair of parentheses is a list of generators of the ideal, all you need to do is write down the products $a_ib_j$, all of them, and see what ideal they generate.

In the present case, we do this: \begin{align} \left(2,1+\sqrt{-5}\,\right)\left(2,1+\sqrt{-5}\,\right)&=\left(4,2+2\sqrt{-5},-4+2\sqrt{-5}\,\right)\\ &=\left(4,2+2\sqrt{-5},2\sqrt{-5}\,\right)\\ &=\left(4,2,2\sqrt{-5}\,\right)\\ &=\left(2,2\sqrt{-5}\,\right)=(2) \end{align} Notice that each simplification is reversible: you can go from left to right, as written, but also from right to left.

Let $\,w = 1\!+\!\sqrt{-5}.\,$ By $\,\rm 4NT =$ Euler's Four Number Theorem and $\,(2,3)\!=\!(1)\,$ we have $$\,\underbrace{2\cdot 3 = \color{0a0}w\!\:\color{c00}{\bar w}\, \Rightarrow\, (2) = (2,\color{#c00}w)(\color{#90f}2,\color{#0a0}{\bar w})}_{\textstyle a\,d\, =\, b\,c\ \underset{\small \rm 4NT}\Rightarrow\, (a) = (a,b)\,(a,c)} = (2,\color{#c00}w)^2\ \ {\rm by}\ \ \underbrace{\color{#0a0}{\bar w}\equiv -\color{#c00}w\!\!\!\pmod{\!\color{#90f}2}}_{\textstyle \bar w + w \,=\, 2}\qquad\qquad$$

As for integers, $\rm 4NT$ constructs a nontrivial (ideal) factorization of $\,2\,$ from a factorization of any multiple $\,w\bar w\,$ of $\,2\,$ that is nontrivial (i.e. $\,2\nmid w,\bar w),\,$ i.e. a witness that $\,2\,$ is composite (not prime). To do so we simply take the ideal gcd (= ideal sum) of $\,2\,$ with each factor $\:\!w,\:\!\bar w\,$ (assuming any common factor of $\color{#0a0}{a,b,c,d}$ has been cancelled). The proof is the same as for gcds in the link, i.e.

$$ \begin{align}&(\color{#0a0}{a,b,c,d})\!=\!(1),\ \color{#c00}{ad = bc} \ \Rightarrow\ (a,b)\,(a,c) = (aa,ab,ac,\color{#c00}{bc}) = (\color{#c00}a)(\color{#0a0}{a,b,c,}\color{#c00}d) = (a)\\[.1em] &\rm\small \color{#0a0}{A\!+\!B\!+\!C\!+\!D}\!=\!(1),\, \color{#c00}{AD\!=\!BC}\Rightarrow(A\!+\!B)(A\!+\!C)=A^2\!+\!AB\!+\!AC\!+\!\color{#c00}{BC}= \color{#c00}A(\color{#0a0}{A\!+\!B\!+\!C}\!+\!\color{#c00}D) = A\end{align} \qquad$$

Alternatively, we can repeat the proof of $\rm 4NT$ in this special case as below, but doing that obfuscates the key arithmetical idea (factorization refinement) that $\rm 4NT$ conveys.

$w^2 =\, 2w\,-\,6\,$ for $\,w = 1+\sqrt{-5},\,$ so apply below with $\,a=2$

$\!\begin{align}w^2 = ab w + ac,\ \color{#c00}{(a,c)\!=\!(1)} \Rightarrow\ (a,w)^2\! &= (a^2,aw,w^2)\\ &=\, (a^2,aw,ac+abw)\\ &=\, a(a,\ \,\color{#0a0}w,\ \,c\:\!+\ \,b\color{#0a0}w)\\ &=\, a(\color{#c00}a,\ \ w,\ \,\color{#c00}c)\\ &=\, a(\color{#c00}1) \end{align}$

The ideal arithmetic in Lubin's answer is a special case of the above. For an introduction to this simple ideal (and gcd) arithmetic see here and its links.