After playing a round of pool billiard recently, I noticed that the fifteen colored balls can be arranged in a triangle (obviously) and all sixteen balls (including the white ball) can be arranged in a perfect square. I asked myself: Are there infinitely many square numbers which are one more than a triangular number? Or put formally: Does the equation $$ \frac{x^2-x}{2}+1=y^2 $$ have infinitely many solutions $(x, y)$, where $x$ and $y$ are integers? I tried to find an answer online, but couldn't find one. Has someone come across this problem before? I am very interested in a proof.
PS: One could generalize this to: For any n, are there infinitely many n-gon-numbers which are one more than a n-1-gon-number?
If $T_n=n(n+1)/2$ is the $n$-th triangular number, then $8T_n+1=(2n+1)^2$ is the square of an odd number. If $T_n=m^2-1$, then $(2n+1)^2=8m^2-7$. Thus we get a solution of $$x^2-8y^2=-7\tag1$$ for positive integers $x$ and $y$. This is a variant of Pell's equation. If $(x_1,y_1)$ is a solution of (1) then so is $(x_2,y_2)=(3x_1+8y_1,x_2+3y_2)$. From the solution $(1,1)$ of (1) we derive further solutions $(11,4)$, $(65,23)$ etc. These give triangular numbers $15$, $528,\ldots$ which are one less than squares.