First question is a well known problem studied by Euler and some variants of it still out of our reach . For every positive integers $(x,y,z)$ let $P(x,y,z)$ denote the predicate: $(xy+1)(yz+1)(zx+1)$ is a square and not all of $(xy+1)$, $(yz+1)$,$(zx+1)$ are squares
Proof by infinite descent:
we well prove by the infinite descent that if there exists a solution $(x,y,z)$ such that $P(x,y,z)$ is true than there exists another solution $(p,q,r)$ such that $P(p,q,r)$ is true and $p+q+r$ less than $x+y+z$.
Let $(x,y,z)$ be tuple of positive integers such that $P(x,y,z)$ is true and $x\leq y \leq z$, consider the two positive integers $s_{+}$ and $s_{-}$ defined by :
$$s_{\mp}=x+y+z+2xyz\mp\sqrt{(xy+1)(yz+1)(zx+1)} \tag 1$$
This integers verify :
$$x^2+y^2+z^2+s_{\mp}^2-2(xy+yz+zx+xs_{\mp}+ys_{\mp}+zs_{\mp})-4xyzs_{\mp}-4=0 \tag2$$
And we can also prove the following important identities (they are basically the same):
$$
\begin{align}
(x+y-z-s_{\mp})^2&&=&& 4(xy+1)(zs_{\mp}+1) \tag 3 \\
(x+z-y-s_{\mp})^2&&=&& 4(xz+1)(ys_{\mp}+1) \tag 4 \\
(x+s_{\mp}-z-y)^2&&=&& 4(yz+1)(xs_{\mp}+1) \tag 5
\end{align}$$
we can use this identities to proof that $P(x,y,s_{\mp})$ holds, by multiplying $(4)$ and $(5)$ you will get that $(xy+1)(ys_{\mp}+1)(xs_{\mp}+1) $ is a square and not all of $(xy+1)$,$(xs_{\mp}+1)$ , $(ys_{\mp}+1)$ are square (as we have $(xz+1)$ is a square iff $xs_{\mp}+1)$ is a square and $(yz+1)$ is a square iff $(ys_{\mp}+1)$ is a square using $(4)$ and $(5)$).
Now the important par is to prove that either $x+y+s_{+}<x+y+z$ or $x+y+s_{-}<x+y+z $ this is equivalent tp proving that $s_{-}s_{+}<z^2$ which is true because :
$$s_{-}s_{+}=x^2+y^2+z^2-2(xy+yz)-4=z^2-x(2z-x)-y(2z-y)-4<z^2 $$
(remember that $x\leq y \leq z$)
Reference:
When Is (xy + 1)(yz + 1)(zx + 1) a Square?
Kiran S. Kedlaya
Mathematics Magazine
Vol. 71, No. 1 (Feb., 1998), pp. 61-63
Second Question:
The following Pell equation have infinitely many solutions :
$$x^2-3y^2=1 $$
For any solution $(n,m)$ of this Pell equation, let $(a,b,c)=(2n-m,2n,2n+m)$ then $ab+1,bc+1,ca+1$ are all squares.
Best Answer
The quick version is $n_0 = 0, \; \; n_1 = 40,$ then $$ \color{magenta}{ n_{k+2} = 98 n_{k+1} - n_k + 40}. $$
Given an $(x,y)$ pair with $3x^2 - 2 y^2 = 1$ we then take $n = (x^2-1)/ 2 = (y^2 - 1)/ 3. $
The first few $x,y$ pairs are $$ x=1, \; y= 1 , \; n=0 $$ $$ x=9, \; y=11, \; n=40 $$ $$ x= 89, \; y=109, \; n=3960 $$ $$ x=881, \; y=1079, \; n= 388080 $$ $$ x=8721, \; y=10681, \; n= 38027920 $$ $$ x=86329, \; y=105731, \; n= 3726348120 $$ and these continue forever with $$ x_{k+2} = 10 x_{k+1} - x_k, $$ $$ y_{k+2} = 10 y_{k+1} - y_k. $$ $$ n_{k+2} = 98 n_{k+1} - n_k + 40. $$
People seem to like these recurrences in one variable. The underlying two-variable recurrence in the pair $(x,y)$ can be abbreviated as $$ (x,y) \; \; \rightarrow \; \; (5x+4y,6x+5y) $$ beginning with $$ (x,y) = (1,1) $$ The two-term recurrences for $x$ and $y$ are just Cayley-Hamilton applied to the matrix $$ A \; = \; \left( \begin{array}{rr} 5 & 4 \\ 6 & 5 \end{array} \right) , $$ that being $$ A^2 - 10 A + I = 0. $$