How did Euler realize $x^4-4x^3+2x^2+4x+4=(x^2-(2+\alpha)x+1+\sqrt{7}+\alpha)(x^2-(2-\alpha)x+1+\sqrt{7}-\alpha)$? How did Euler find this factorization?
$$\small x^4 −  4x^3 + 2x^2 + 4x + 4=(x^2-(2+\alpha)x+1+\sqrt{7}+\alpha)(x^2-(2-\alpha)x+1+\sqrt{7}-\alpha)$$
where $\alpha = \sqrt{4+2\sqrt{7}}$
I know that he had some super powers, like he was sent to us from a super intelligent alien universe just to humiliate our intelligence, but how the hell did he do that three centuries ago? :|

The other answers are great but were pure speculation since Euler published how he solves quartics.

From Elements of Algebra by Euler section 4 chapter 15, his new method for resolving fourth order equations. Additionally, Google books has a full copy to download or read online here.
Suppose the root of the equation is of the form $x = \sqrt{p} + \sqrt{q} + \sqrt{r}$ where $p,q,r$ are the roots of an equation of degree three,
$$
z^3 -fz^2+gz-h=0
$$
and 
\begin{align}
f&=p+q+r\tag{1}\\
g&=pq+pr+qr\tag{2}\\
h&=pqr\tag{3}
\end{align}
Now, he square $x$ and obtained
$$
x^2 = p+q+r+2\sqrt{pq}+2\sqrt{pr}+2\sqrt{qr}
$$
From $(1)$, we can now write
$$
x^2-f=2\sqrt{pq}+2\sqrt{pr}+2\sqrt{qr}\tag{4}
$$
Let's square $(4)$ again to obtain:
$$
x^4-2fx^2+f^2 = 4pq+4pr+4qr+8\sqrt{p^2qr}+8\sqrt{pq^2r}+8\sqrt{pqr^2}
$$
From $(2)$, we can now write
$$
x^4-2fx^2+f^2 -4g= 8\sqrt{pqr}(\sqrt{p}+\sqrt{q}+\sqrt{r})
$$
Using the identity first laid out for $x$ and $(3)$, we have
$$
x^4-2fx^2 -8x\sqrt{h}+f^2-4g= 0\tag{5}
$$
Euler says we don't need to worry about $yx^3$ because

for we shall afterwards shew, that every complete equation may be transformed into another, from which the second term is taken away.

Next, Euler goes onto factoring a quartic as a product of two real quadratics.

It is to be observed that the product of these three terms, or $\sqrt{pqr}$, must be equal to $\sqrt{h}=b/8$, and that if $b/8$ is positive, the product of the terms $\sqrt{p},\sqrt{q},\sqrt{r}$; must likewise be positive so that all variations that can be admitted are reduced to the four following:

\begin{align}
x &= \sqrt{p} + \sqrt{q} + \sqrt{r}\\
x &= \sqrt{p} - \sqrt{q} - \sqrt{r}\\
x &= -\sqrt{p} + \sqrt{q} - \sqrt{r}\\
x &= -\sqrt{p} - \sqrt{q} + \sqrt{r}
\end{align}
When $b/8$ is negative, we have
\begin{align}
x &= \sqrt{p} + \sqrt{q} - \sqrt{r}\\
x &= \sqrt{p} - \sqrt{q} + \sqrt{r}\\
x &= -\sqrt{p} + \sqrt{q} + \sqrt{r}\\
x &= -\sqrt{p} - \sqrt{q} - \sqrt{r}
\end{align}

In order to use Euler's method, you would first need to transform your equation such that the $x^3$ term is gone. Then you match up the coefficients of $f,g,h$ in your new polynomial with the general form $(5)$. That is, given a polynomial of the form
$$
x^4 - ax^2 - bx - c = 0
$$
you would set $a = 2f$, $b = 8\sqrt{h}$, and $-c = f^2 - 4g$. Then you can follow his book's examples which you will find in the second hyperlink.
