I tried solving this so first $1^2+2^2+\cdots+n^2=n(n+1)(2n+1)/6$ so problem is $n(n+1)(2n+1)=6y^2$ where $n$ and $y$ are natural. I tried doing this $n=6k,6k+1....6k+5$, so say for $n=6k$
you get $k(6k+1)(12k+1)=y^2$ so $k(6k+1)$ and $12k+1$ are co-prime which means they are all squares meaning $k=a^2 \quad 6k+1=b^2\quad 12k+1=c^2$ but idk what to do from here?
Whatever I try I can't solve this.
ps. I'm pretty sure $n=24, y=70$ is the only solution besides $1,1$.