Given $k$, can we find $n$ such that $d$ is a perfect square

Question

Given an odd positive integer $k$. Define $$d=16(k+1)^3(k+2)(n+1)^2+1$$ where $n$ is alos a positive integer.

My question is: Given $k$, can we find $n$ such that $d$ is a perfect square.

Answer 1

$$d=16(k+1)^3(k+2)(n+1)^2+1\tag{1}$$ Let $d=x^2, y=4(k+1)(n+1), N=(k+1)(k+2)$ then we get $x^2-Ny^2=1.$
This is a Pell equation.
According to Pell equation wiki, all solutions of $x^2-Ny^2=1$ are given as follows.

$x_m+y_m\sqrt{N}=(x_1+y_1\sqrt{N})^m$

For instance, let $k=1$ then we get $x^2-6y^2=1.$
We get the solutions from the obvious solution $(x_1,y_1)=(5,2)$ with $m=1..9$ as follows.

            (k,x,y)
            (1,5,2)
            (1,49,20)
            (1,485,198)
            (1,4801,1960)
            (1,47525,19402)
            (1,470449,192060)
            (1,4656965,1901198)
            (1,46099201,18819920)
            (1,456335045,186298002)

Hence we get the solution $(x,y)=(4801,1960) \implies (k,d,n)=(1, 4801, 244).$
Similarly, we get $(k,d,n)=(1, 46099201, 2352489).$

Other solutions(smallest solution) are as follows.
These solutions have already been found by Dmitry Ezhov and Will Jagy.

$(k,d,n)$
$(3, 5374978561, 75117608)$
$(5, 46873096812360001, 301361494218524)$
$(7, 1572584048032918633353217, 5791587730988896551760)$
$(9, 144220715637070429940775452568001, 343772642385433639988435123780)$