How to prove that 2 is not a Gaussian prime? To prove an algebraic number is not a Gaussian prime, need find factors of it.
Let us assume, $(a + bi)(x +yi) = 2, \exists a,b,x,y \in \mathbb{Z}$; so $(ax -by) + i(ay + bx) =2$.
As imaginary part is null in $2$, so $ax - by =2, ay + bx =0$.
Now, how to reach from here, the fact that $a=x=1, b=i, y=-i$.
I read in here in a comment by @StevenStadnicki, that an easier way is to multiply both sides of $(a + bi)(x +yi) = 2$ by the complex conjugates, but is not clear, as my below attempt shows.
$$
\begin{align}
& ax - by =2, ay + bx =0  & \ \\
\implies & axy - by^2 =2y, axy + bx^2 =0 \text{ }{ -(i)} & \ \\
\implies & b = 0
\end{align}
$$

By the answer given below, my attempt is stated below:
Two set of values are possible, one in which $(a^2 + b^2) = 4, (x^2 + y^2) = 1$, second in which $(a^2 + b^2) = 2, (x^2 + y^2) = 2$.
Possible combinations for each of the two cases are stated below:
Case (i): $(a^2 + b^2) = 4, (x^2 + y^2) = 1$ $\implies  a= \pm 2, b = 0, x = \pm 1, y = 0$, or $a= \pm 2, b = 0, x = 0, y = \pm 1$.  But, as $ax -by =2, ay + bx = 0$.  But being algebraic number, need have non-zero imaginary part, hence no algebraic factors of $2$.
Case (ii)  $(a^2 + b^2) = 2, (x^2 + y^2) = 2$ $\implies$ possible values are : $  a= \pm 1, b = \pm 1, x = \pm 1, y = \pm 1$. 
But, as : $ax -by =2, ay + bx = 0$, so the valid combinations are: 1. $a=1, b=-1, x=1, y=1$; 2. $a=1, b=1, x=1, y=-1$, 3. $a=-1, b=1, x=-1, y=-1$. 
Here, combination 3 is  same as 1, so can be ignored  also valid.
So, the solution has values of $a =1,b=1, x = 1, y = -1$.
If there is a redundancy or unnecessary steps, please tell.

If $(a+bi)(x+yi)=2$, $(a-bi)(x-yi)=2$ and therefore $(a^2+b^2)(x^2+y^2)=4$. In how many ways can $4$ be written as the product of two integers, each of which is a sum of $2$ squares? Well, $4=4\times1=(2^2+0^2)\times(1^2+0^2)$ and $4=2\times2=(1^2+1^2)\times(1^2+1^2)$. Therefore, there are few possibilities for you to test.
