On the extension of $D(-8k^2)$-pair $\{8k^2,8k^2+1\}$

Let $k$ be a positive integer. The triple $\{1,8k^2,8k^2+1\}$ has the property that the product of any two of its distinct elements subtracted by $8k^2$ is a perfect square. By elementary means, we show that this triple can be extended to at most a quadruple retaining this property, i.e., if $\{1,8k^2,8k^2+1,d\}$ has the same property, then $d$ is uniquely determined ($d=32k^2+1$). Moreover, we show that even the pair $\{8k^2,8k^2+1\}$ can be extended in the same manner to at most a quadruple (the third and fourth element can only be $1$ and $32k^2+1$). At the end, we suggest considering a similar problem of extending the triple $\{1,2k^2,2k^2+2k+1\}$ with a similar property as possible future research direction.