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.
Best Answer
This rather elliptic argument (for twenty-five) is really mostly adapting Euler’s proof, which is extremely accessible (there’s no denying a high-schooler studying for the IMO could do that given enough time and perhaps a little encouragement). I admit not being too sure how to consider the case of less integers without redoing some 24 times the calculations.
Consider twenty-five consecutive integers the product of which is a square.
Define, for an integer $n$, the radical of $n$ to be the product of the prime divisors of $n$ who appear an odd number of times in the prime decomposition of $n$. We’ll consider the radicals of our $25$ consecutive integers (so they are square-free, and products of primrs less than $25$).
So, with elementary modular arithmetic, one finds that two appears in seven to ten radicals. Three appears in five to seven radicals. Five appears in four to five radicals. Seven appears in two to four radicals. Eleven appears in one to three radicals. Primes above eleven appear in at most two radicals.
Because the product of the consecutive integers is a square, two appears in eight or ten radicals, three in six, five in four, seven in two or four, eleven in two, primes above eleven in zero or two.
It follows that the product of the radicals divides $N=2^{10}3^65^47^4(11\cdot 13 \cdot 17 \cdot 19 \cdot 23)^2$.
But what are the first $25$ square-free numbers? They are $1,2,3,5,6,7,10,11,13,14,15,17,19,21,22,23,26,29,30,31,33,34,35,37,38$. Their product is $M=2^93^65^57^411^313^217^219^2 \cdot 23 \cdot (29 \cdot 37)$ so $M > N$.
It follows from this that two of the 25 numbers (let’s call them $a < b$) have the same radical $r$, and thus we can write $a=rx^2,b=ry^2$ and $r(y^2-x^2) \leq 24$. So $r(2y+1) \leq 24$, thus $ry \leq 12$, thus $b \leq (ry)^2 \leq 144$. We can check (counting powers of $19$ and $23$) that this is impossible.