Hint: Let $p=2k+1$ where $k \in \mathbb{N},$ then $2^{2k}-1=(2^k-1)(2^k+1)=pm^2.$ We know that $\gcd(2^k-1,2^k+1)=1$ since they are consecutive odd integers, so the equation breaks into $2^k-1=px^2, 2^k+1=y^2$ or $2^k-1=x^2, 2^k+1=py^2.$
Easy investigation shows that the only solutions are $p=3,7.$ I will leave it to you to fill the gaps. You may also interested in this.
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
If $p^3-4p+9=n^2$ then $n^2\equiv 9 \pmod{p}$ and $\pm n \equiv 3 \pmod{p}$ since $n^2-9\equiv 0$ can only have two roots modulo a prime.
By choosing the appropriate sign we can write $$ \begin{align} n^2=(-n)^2 &= (kp+3)^2 \\ p^3-4p+9 &= k^2p^2+6kp+9 \\ 0 &= p^2-k^2p-(6k+4) \\ p & = \frac{k^2\pm \sqrt{k^4+24k+16}}{2} \end{align} $$ where we used $p\ne 0$ since it's a prime. This can only have an integer solution for $p$ when $k^4+24k+16$ is a square (as @tkr suggested in the comments). But $$ \begin{array}{ll} (k^2-1)^2 = k^4-2k^2+1 < k^4+24k+16<(k^2)^2 & \text{if }k<-11 \\ (k^2)^2 < k^4+24k+16< k^4+2k^2+1 =(k^2+1)^2 & \text{if }k>12 \end{array} $$ so if $k<-11$ or $k>12$ then $k^4+24k+16$ lies between two squares and hence is not a square. This leaves 24 values to check, which you could do by hand. It leads to solutions when $k=-3,0,3$ which give integral values for $p$ of $-2,2,7,11$, of which only $2,7,11$ would normally be considered prime.