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
I assume the arithmetic progression is non-degenerate, i.e., $d \neq 0$. Also, since the progression is of positive integers, then $a \gt 0$ and $d \gt 0$.
Let $a + kd$, for some integer $k \ge 0$, be the smallest element of the progression which is a perfect square. Thus, there's a positive integer $e$ such that
$$a + kd = e^2 \tag{1}\label{eq1A}$$
You're asking to prove
$$e^2 \lt a + 2d\sqrt{a} + d^2 \tag{2}\label{eq2A}$$
As Joffan's question comment states, the right side is equivalent to $(\sqrt{a} + d)^2$. Thus, \eqref{eq2A} is equivalent to trying to prove
$$e \lt \sqrt{a} + d \tag{3}\label{eq3A}$$
Assume, instead, that
$$e \ge \sqrt{a} + d \implies e - d \ge \sqrt{a} \implies (e - d)^2 \ge a \tag{4}\label{eq4A}$$
Note, though, that
$$\begin{equation}\begin{aligned} (e - d)^2 & = e^2 - 2ed + d^2 \\ & = (a + kd) - 2ed + d^2 \\ & = a + (k - 2e + d)d \end{aligned}\end{equation}\tag{5}\label{eq5A}$$
This is also a perfect square smaller than $e^2$ (due to $e \gt e - d \gt 0$ using \eqref{eq4A}), and of the form $a + jd$ for an integer $j = k - 2e + d$, so it's a member of the arithmetic progression if $j \ge 0$. However, since $e^2$ is the smallest perfect square which is a member of the progression, this means $(e - d)^2$ cannot be a member, so this requires $j \lt 0$ and
$$(e - d)^2 \lt a \tag{6}\label{eq6A}$$
Note this contradicts \eqref{eq4A}, so the assumption it is correct is false. This means \eqref{eq3A} and, thus, \eqref{eq2A}, must be true instead.