product of six consecutive integers being a perfect square

I am not sure how elementary you want your proof to be, but here is a proof that uses elliptic curves...

Suppose that there are $x,y\in\mathbb{Z}$ such that $x>0$ and

$$y^2=x(x+1)(x+2)(x+3)(x+4)(x+5).$$

If we put $t=x+2+\frac{1}{2}$, then we have

$$y^2=(t-5/2)(t-3/2)(t-1/2)(t+1/2)(t+3/2)(t+5/2)=\left(t^2-\frac{1}{4}\right)\left(t^2-\frac{9}{4}\right)\left(t^2-\frac{25}{4}\right),$$

or, equivalently,

$$4^3y^2 = (4t^2-1)(4t^2-9)(4t^2-25).$$

If we put $U=2^3y$ and $V=4t^2$, then we have a solution for the equation

$$U^2=(V-1)(V-9)(V-25)=V^3 - 35V^2 + 259V - 225.$$

This defines an elliptic curve $E/\mathbb{Q}$, and we can use standard techniques to calculate the rank of the group of rational points $E(\mathbb{Q})$. This method ($2$-descent) shows that the rank of the curve is $0$, and one can easily separately show that the torsion subgroup is $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$. It follows that the only points on $E(\mathbb{Q})$ are the trivial points $(V,U)=(1,0)$, $(9,0)$ and $(25,0)$, plus the point ``at infinity'' on the curve. These correspond to $t$-values $t=\pm 1/2$, $\pm 3/2$ and $\pm 5/2$, and therefore do not give any integer values of $x$ with $x> 0$. Hence, there are no integer solutions to our original equation.

There is probably some elementary argument that shows that $U^2=(V-1)(V-9)(V-25)$ only has $3$ solutions, but I can't think of one right away.

It's possible to adapt Ross Millikan's argument :

Assuming none of the numbers is $0$, since primes greater than $5$ can only appear once, each of the six numbers is of the form $2^a 3^b 5^c y^2$ with $a,b,c = 0$ or $1$. Also, each prime $2,3,5$ can only appear an even number of time in order for the product to be a square.

If two of the six numbers have the same exponent triple $(a,b,c)$, it puts a very small upper bound on $x$ because it implies that we have two squares $y^2$ extremely close together. So the goal is to try to give them six different exponent triples out of the $8$ available and fail.

If the prime $5$ doesn't appear, you only have $4$ triples for $6$ numbers, impossible. So $5$ has to appear twice in $x$ and $x+5$ only.

Then the four numbers in the middle have to take the other $4$ triples with no $5$, so the prime $3$ must appear in $x+1$ and $x+4$ only.

Finally, the prime $2$ has to appear twice in the middle numbers only. It can't appear three times or else the whole product wouldn't be a square, and it can't appear four times because there is not enough room.

Thus the numbers $x$ and $x+5$ must be of the form $5y^2$. So whatever we do we get an upper bound on $x$.