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.
Yeah, there's infinitely many. You can see this by considering a particular case:
if $3n$ and $n+1$ are both squares.
we can parametrize this by $n = 3k^2$.
Then we need $3k^2+1=a^2$
So we want to solve $a^2-3k^2=1$ which is a Pell equation.
the fundamental solution $(a_1,k_1)$ is $(2,1)$ and the other solutions are obtained via the recurrence $a_{n+1}=2a_n + 3k_n, k_{n+1} = a_n+2k_n$.
So we get the solutions are:
$(7,4),(26,15),(97,56), \dots$
Best Answer
Since $n\le 668$, we know that $7n+1\le 4677$. $\left\lfloor\sqrt{4677}\right\rfloor=68$, so we want to know how many of the integers $1,2,\ldots,68$ have squares of the form $7n+1$. Suppose that $7n+1=k^2$ for some integer $k$; then $k^2\equiv 1\pmod 7$, and it’s easy to check that this is the case precisely when $k\equiv 1\pmod 7$ or $k\equiv -1\pmod 7$, i.e., when $k=7m\pm 1$ for some integer $m$.
There are $9$ positive multiples of $7$ less than $68$, each of which provides two such values of $k$.
Thus, there are altogether $2\cdot9=18$ such integers $n$.