Find all $x$ so $\frac{1^2+2^2+\cdots+x^2}{x}$ is a perfect square... Find all $x$ so $\dfrac{1^2+2^2+\cdots+x^2}{x}$ is a perfect square...
Clearly 1 is solution, but I have to show that there is an infinity... I see that this happens when the numbers $1,2,...,x$ have a whole quadratic root. Am I on the right road?

The main result is:
$$1^2+2^2+3^2+\dots + x^2=\frac{x(x+1)(2x+1)}{6}$$
So you need $\frac{(x+1)(2x+1)}{6}=y^2$ for some $y$. 
Or:
$$2x^2+3x+1 = 6y^2$$
From here, we are essentially going to "complete the square" on the left side.
Multiply both sides by $8$ and you get:
$$16x^2+24x+8=(4x+3)^2-1=48y^2$$
So you need infinitely many solutions to:
$$(4x+3)^2-48y^2=1$$
So you need to know how to solve Pell's equations from here.
The generally positive solution to the equation $u^2-48y^2=1$ can be written as:
$$u+y\sqrt{48}=(7+\sqrt{48})^k$$
For any positive integer $k$. 
You should be able to prove with induction that infinitely many $u$ have $u\equiv 3\pmod 4$.
The recursive sequence of values of $u$ is $u_0=1,u_1=7$ and $u_{n+1}=14u_n-u_{n-1}$. You can show that $u_k\equiv 3\pmod 4$ when $k$ is odd. The next answers after $x=1$ are $x=337$ and $x=65521$. 
