Proving the product of four consecutive integers, plus one, is a square I need some help with a Proof:

Let $m\in\mathbb{Z}$. Prove that if $m$ is the product of four consecutive integers, then $m+1$ is a perfect square.

I tried a direct proof where I said:
Assume $m$ is the product of four consecutive integers.
If $m$ is the product of four consecutive integers, then write $m=x(x+1)(x+2)(x+3)$ where $x$ is an integer.
Then $m=x(x+1)(x+2)(x+3)=x^4+6x^3+11x^2 +6x$.
Adding $1$ to both sides gives us:
$m+1=x^4+6x^3+11x^2+6x+1$.
I'm unsure how to proceed. I know I'm supposed to show $m$ is a perfect square, so I should somehow show that $m+1=a^2$ for some $a\in\mathbb{Z}$, but at this point, I can't alter the right hand side of the equation to get anything viable.

If the numbers are $x, x+1, x+2, x+3$. 
Let $\frac m2 = x+1.5$ be the midpoint of the four consecutive integers, so that the integers are $\frac {m-3}2, \frac {m-1}2, \frac {m+1}2, \frac {m+3}2$.  (Note: $m$ is odd and $\frac m2$ is not an integer.)
So $x(x+1)(x+2)(x+3) + 1 = $
$\frac {(m-3)(m+3)(m-1)(m+1)}{16} + 1=$
$\frac {(m^2 -9)(m^2 - 1) + 16}{16} =$
$\frac {(m^2-10m^2 + 9) +16}{16} =\frac {m^2-10m^2 + 25}{16}=$
$(\frac {m^2 -5}{4})^2$.
Now $m$ is odd.  So let $m = 2n+1$ then 
$x(x+1)(x+2)(x+3) + 1 = (\frac {(2n+1)^2 -5}{4})^2=$
$(\frac {4n^2 +4n + 1 -5}{4})^2 = (\frac {4n^2 +4n -4}{4})^2=$
$(n^2 +n - 1)^2$.
====
So ....
If $x$ is the first integer and $\frac {2n + 1}2 = x + \frac 32$ then
$n = x + 1$.
So $x(x+1)(x+2)(x+3) + 1 = ((x+1)^2 + (x+1) - 1)^2 = (x^2 + 3x +1)^2$
.... 
So as you got $x^4 + 6x^3 + 11x^2 + 6x + 1$ that actually equals $(x^2 + 3x +1)^2$
Indeed $x^4 + 6x^3 + 11x^2 + 6x + 1 = x^2(x^2 + 3x + 1) + 3x^3 +10x^2 + 6x + 1$
$=  x^2(x^2 + 3x+ 1) + 3x(x^2 + 3x + 1) +x^2 +3x + 1$
$= (x^2 + 3x + 1)^2$.
.... addendum.....
D'oh.
If $x(x+1)(x+2)(x+3) + 1 = a^2$ then 
$x(x+ 1)(x+ 2)(x+3) = a^2 - 1 = (a + 1)(a-1)$
To get factors that close together they'd have to be 
$a = x(x+3) \pm 1= (x+ 1)(x+2) \mp 1$
and indeed $a= x(x+3) + 1 = (x+1)(x+2) - 1= x^2 +3x + 1$.
proves the statement!  (If we work backwards.)
