Integer solutions to the product of four consecutive integers $ \bullet \textbf{Question} $
The product of four consecutive integers $ x, x + 1, x + 2, x + 3 $ can be written as the product of two consecutive integers, find all integer solutions for $ x $.
$ \bullet \textbf{Rephrasing} $
I decided to name the other integer as $ y $ so that this equation is a plot on a graph,
$$ y(y + 1) = x(x + 1)(x + 2)(x + 3) \tag{1} $$
$ \bullet \textbf{Attempt} $
To solve for $ y $ I did these simple steps,
$$ y^2 + y = x(x + 1)(x + 2)(x + 3) \tag{2} $$
$$ y^2 + y - x(x + 1)(x + 2)(x + 3) = 0 \tag{3} $$
$$ y = \frac{-1 \pm{\sqrt{1 + 4x(x + 1)(x + 2)(x + 3)}}}{2} \tag{4} $$
I then figured that when,
$$ -1 \pm{\sqrt{1 + 4x(x + 1)(x + 2)(x + 3)}} \equiv 0\pmod{2} \tag{5} $$ 
then $ y $ will be a integer, so this is just solving for all $ x $, so $ y $ will be a integer but not for only integer $ x $'s.
This then means that 
$$ \pm{\sqrt{1 + 4x(x + 1)(x + 2)(x + 3)}} \equiv 1\pmod{2} \tag{6} $$
$$ {\sqrt{1 + 4x(x + 1)(x + 2)(x + 3)}} \equiv 1\pmod{2} \tag{7} $$
This is the end of my knowledge, if you try to square it will include the integer solution for $ y $ but also include non-integer solutions for $ y $.
Am I going in the correct direction, or is this a dead end and I need to apply a new approach?

We have $$y^2+y=(x^2+3x)(x^2+3x+2)$$ or$$y^2+y+1=(x^2+3x+1)^2$$ or
$$(2y+1)^2+3=(2x^2+6x+2)^2$$ or
$$(2x^2+6x-2y+1)(2x^2+6x+2y+3)=3.$$
Now, solve a number of systems.
