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.
In the case of $14444$ base $b\ge5$ there can be no squares. We have
$14400_b<14444_b<14641_b$
$(b^2+2b)^2<14444_b<(b^2+2b+1)^2$
and the left and right members are consecutive squares.
With six 4's again there must be no squares:
$1442401_b<1444444_b<1444804_b$
$(b^3+2b^2+1)^2<1444444_b<(b^3+2b^2+2)^2$
We can go on like this. Consider the Laurent series
$\sqrt{1.4444..._b}=\sqrt{1+\dfrac{4b^{-1}}{1-b^{-1}}}=1+\sum\limits_{k=1}^\infty a_kb^{-k}$
$a_1=2,a_2=0,a_3=2,a_4=-
2,...;a_k\in\mathbb{Z}\text{ for all } k$
If $a_m$ is followed by $m$ zeroes in the coefficient sequence then
$P(m)=b^m\left(1+\sum\limits_{k=1}^m a_kb^{-k}\right)$
will be an exact square root of $1444..44_b$ with $n=2m$ fours in all bases; but this can happen only for $m\in\{0,1\}$ because the the known limitations in base ten. For larger $m$, $P(m)$ becomes one of two consecutive whole numbers that strictly bracket $\sqrt{1444...44_b}$. For example, $P(3)=b^3+2b+2$ and we saw above that for $n=2×3=6$ the target square root lies strictly between $P(3)-1$ and $P(3)$. Thereby there are no squares in any base with an even number $\ge4$ of fours.
Best Answer
In general, the Diophantine equation $$ n!+k=m^2 $$ is "easy" for $k$ being a non-square integer and hard for $k$ being a perfect square. In fact, if $k$ is not a square it follows that $n\le k$. In this case, for small $k$, we only have to check a few possible values for $n$. The most famous case, where $k$ is a perfect square is the Brocard equation $n!+1=m^2$, which is unsolved until today.
For a reference see the following post with very informative answers:
For any $k \gt 1$, if $n!+k$ is a square then will $n \le k$ always be true?