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.

Given $m$ is the product of four consecutive integers.
$$m=p(p+1)(p+2)(p+3)$$where $p$ is an integer
we need to show that $p(p+1)(p+2)(p+3)+1$ is a perfect square 
Now,$$p(p+1)(p+2)(p+3)+1=p(p+3)(p+1)(p+2)+1$$
$$=(p^2+3p)(p^2+3p+2)+1$$
$$=(p^2+3p+1)(p^2+3p+2)-(p^2+3p+2)+1$$
$$=(p^2+3p+1)(p^2+3p+1+1)-(p^2+3p+2)+1$$
$$=(p^2+3p+1)(p^2+3p+1)+(p^2+3p+1)-p^2-3p-2+1$$
$$=(p^2+3p+1)(p^2+3p+1)=(p^2+3p+1)^2$$
So, $m+1$ is a perfect square where $m$ is the product of four consecutive integers.
