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.

This is another way of looking at @jwg's answer.

Let the four consecutive numbers be $a,b,c,d$ and let $t$ be the number half-way between $b$ and $c$. Then 
\begin{align}
   abcd + 1
   &= \bigg(t-\frac 32\bigg)\bigg(t-\frac 12\bigg)
      \bigg(t+\frac 12\bigg)\bigg(t+\frac 32\bigg) + 1\\
   &= \bigg(t^2-\frac 94\bigg)\bigg(t^2-\frac 14\bigg) + 1\\
   &= t^4 - \frac 52 t + \frac{25}{16}\\
   &= \bigg(t^2 - \frac 54 \bigg)^2
\end{align}
Letting $t = a + \frac 32$, we get
\begin{align}
   abcd + 1
   &= \bigg(t^2 - \frac 54 \bigg)^2 \\
   &= \bigg(a^2 +3a + \dfrac 94 - \frac 54 \bigg)^2 \\
   &= (a^2 +3a + 1)^2
\end{align}
This suggests the following solution.
\begin{align}
   abcd + 1
   &= a(a+1)(a+2)(a+3) + 1 \\
   &= a(a+3) \cdot (a+1)(a+2) + 1 \\
   &= (a^2+3a) (a^2+3a+2) + 1 \\
   &= (a^2+3a+1 \ - \ 1)(a^2+3a+1 \ + \ 1) + 1 \\
   &= (a^2+3a+1)^2 - 1 + 1 \\
   &= (a^2+3a+1)^2
\end{align}
