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.

Empirically:
Consider the function 
$$p(n):=\sqrt{n(n+1)(n+2)(n+3)+1}.$$
For $n=0,1,2,3,\cdots$ we have $p(n)=1,5,11,19,29,\cdots$ a sequence with constant second order differences ($2$), and we can postulate the polynomial 
$$n^2+3n+1$$
(because $p(0)=1$, the coefficient of $n^2$ must be $1$, and $p(n)-n^2-1=0,3,6,9,\cdots$)
Now the identity
$$n(n+3)(n+1)(n+2)=(n^2+3n)(n^2+3n+2)
\\=(n^2+3n+1-1)(n^2+3n+1+1)
\\=(n^2+3n+1)^2-1.$$
becomes apparent.

Another approach is by bringing more symmetry and shifting the variable by $3/2$.
$$\sqrt{\left(m-\frac32\right)\left(m-\frac12\right)\left(m+\frac12\right)\left(m+\frac32\right)+1}
=\sqrt{\left(m^2-\frac94\right)\left(m^2-\frac14\right)+1}
=\sqrt{m^4-\frac52m^2+\frac{25}{16}}=m^2-\frac54,
$$
which is 
$$n^2+2\frac32n+\left(\frac32\right)^2-\frac54.$$
