This question comes from trying to see why 24 is the only non-trivial value of $n$ for which
$$1^2+2^2+3^2+\cdots+n^2$$
is a perfect square.
To this end, let $m,n \in \mathbb N$ be such that $1^2+2^2+3^2+\cdots+n^2 = m^2$, or
$$n(n+1)(2n+1) = 6m^2.$$
When $n=24$ the left hand side is $24\times 25 \times 49$ and there are two things that make it work as a solution:
$$7^2+1=2\times5^2$$
$$7^2-1=12\times2^2.$$
We can write these algebraically as
$$x^2=2y^2-1$$
$$x^2=12z^2+1$$
and solve them simultaneously (with $x,y,z\in \mathbb N$).
These are instances of Pell's equation and each individually has an infinite number of solutions.
How do we show there is only one (non-trivial) value of $x$ that is common to the solutions of both equations?