In the case of $(5, 442, 541)$, the product of each two of them minus 1 is a square: $$5 \times 442 - 1 = 47^2, 5 \times 541 - 1 = 52^2, 442\times541 - 1 = 489^2$$
What are the integer-solutions $(w, x, y, z)$ for the case of four numbers, i.e. $$w x - 1 = a^2, w y - 1 = b^2, w z - 1 = c^2$$ $$x y - 1 = d^2, x z - 1 = e^2, y z - 1 = f^2$$ where $a, b, c, d, e, f$ are integers?
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I do not think this is possible; I think that two out of four numbers must be even, so that their product is a multiple of $4$ and not one larger than a square.....
Friday: after looking at the articles by Dujella recommended in the accepted answer, it turns out, to get a quadruple, one of the elements must be exactly $1.$ This means that we want $A,B,C$ each a square plus one, and each pairwise product also a square plus one. So I ran a simple program to do that, the numbers $A,B$ both squares plus one, then find all $C$ up to a large bound and see what happens. There is some unpredictable stuff, but the majority of triples come in infinite families, $$ A = t^2 + 1, \; \; B = (t+1)^2 + 1, \; \; C = (2t+1)^2 + 4, $$ $$ A = t^2 + 1, \; \; B = 4 t^4 + 1, \; \; C = 16 t^8 - 4 t^4 + 1, $$ $$ A = t^2 + 1, \; \; B = 4 t^4 + 1, \; \; C = 4 t^4 - 4 t^3 + t^2 - 2t + 2, $$ $$ A = t^2 + 1, \; \; B = 4 t^4 + 1, \; \; C = 4 t^4 + 4 t^3 + t^2 + 2t + 2. $$ The first triple comes very close, $4$ instead of $1.$ The $C$ value in the second triple cannot be a square plus one either. So, as usual, it comes down to the unpredictable triples.
Here is something similar that combines polynomials with a Pell type equation: Take $t$ from the sequence $3, 11, 41, 153, 571, 2131, \ldots$ so that $t_{k+2} = 4 t_{k+1} - t_k,$ with the result that we get an integer $w$ with $w = \sqrt{3t^2 - 2}.$ Then let $$ m = -t + w, \; \; \; n = t + w, $$ with the result that $m,n$ are consecutive terms in the sequence $$ 2, 8, 30, 112, 418, 1560, 5822, 21728, $$ which follow the same linear recursion as $t_k.$
Finally $$ A = m^2 + 1, \; \; B = n^2 + 1, \; \; C = 4 t^2 - 2, $$ $$ A = m^2 + 1, \; \; B = n^2 + 1, \; \; C = 12 t^2 - 2, $$ $$ A = m^2 + 1, \; \; B = n^2 + 1, \; \; C = 48 t^4 + 16 t^2 + 1. $$
Let me prove it with the first two numbers being $5$ and $13,$ with $5 \cdot 13 = 65 = 8^2 + 1.$ We can see that taking the third number to be $2$ works. However, what can we say about $C$ when $5C = V^2 + 1$ and $13C = W^2 + 1.$ Well, $$ 65C = 13 V^2 + 13 = 5W^2 + 5, $$ $$ 5W^2 - 13V^2 = 8. $$ The base solutions with positive entries are $$ W = 5, \; \; V = 3, $$ $$ W = 21, \; \; V = 13. $$ All other solutions are derived from these by negating either $W$ or $V$ and using the transformation $$ (W,V) \mapsto (129W + 208V, 80W + 129V). $$ The inverse of the transformation is just $$ (W,V) \mapsto (129W - 208V, -80W + 129V). $$ That is, all solutions have both $W,V$ odd. Therefore $V^2 + 1$ is even and so is $C.$
I suspect there is a simple algebraic number theory proof of this that works for any initial pair of odd numbers $A,B$ with $AB = U^2 + 1.$ Gaussian Integers or something.
Genaeral shape of it: given two positive odd numbers $A,B$ with $AB = U^2 + 1,$ it is true that we can write $$ A = \alpha^2 + \beta^2, \; \; \; B = \gamma^2 + \delta^2, $$ with $$ \det \left( \begin{array}{rr} \alpha & \beta \\ \gamma & \delta \end{array} \right) \; = \; 1 $$ I already believe that all possible values of the third number $C$ can be expressed as $$ C = (-\alpha + \gamma)^2 + (-\beta + \delta)^2 = A + B - 2 \alpha \gamma - 2 \beta \delta$$ which is even. I say all possibilities because, when $A$ and $B$ have many distinct prime factors, there will be many possible ways to choose $\alpha, \beta, \gamma, \delta,$ so I would expect several different $C.$ However, when $A,B$ are odd, $A+B$ is even and so is any value of $C.$ Needs a careful proof, I do not yet see the whole thing.
EXAMPLE: $A=65,$ $B = 901.$ The only matrix that works is $$ \det \left( \begin{array}{rr} 7 & 4 \\ 26 & 15 \end{array} \right) \; = \; 1 $$ $$ \det \left( \begin{array}{rr} 7 & 4 \\ 19 & 11 \end{array} \right) \; = \; 1 $$ $$ \det \left( \begin{array}{rr} 26 & 15 \\ 19 & 11 \end{array} \right) \; = \; 1 $$ and $C= 19^2 + 11^2 = 482$
SMALL STEPS Wednesday 9 March: with $AB=U^2 + 1,$ once $A,B > 1$ and $A \neq B,$ the recipe above will always give $$ C = A+B-2U. $$ Perhaps there really is only one value of $C?$ If so, the answer to the OP is a simple no.
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Nicolae Ciprian Bonciocat, Mihai Cipu, and Maurice Mignotte, There is no Diophantine $D(-1)$-quadruple, Journal of the London Mathematical Society, Volume 105, Issue 1, January 2022, Pages 63-99, https://doi.org/10.1112/jlms.12507 prove that there are no four distinct positive integers such that each pair has product one more than a square. Here is the abstract:
A set of positive integers with the property that the product of any two of them is the successor of a perfect square is called Diophantine $D(-1)$-set. Such objects are usually studied via a system of generalized Pell equations naturally attached to the set under scrutiny. In this paper, an innovative technique is introduced in the study of Diophantine $D(-1)$-quadruples. The main novelty is the uncovering of a quadratic equation relating various parameters describing a hypothetical $D(-1)$-quadruple with integer entries. In combination with extensive computations, this idea leads to the confirmation of the conjecture according to which there is no Diophantine $D(-1)$-quadruples.
Currently, access at https://londmathsoc.onlinelibrary.wiley.com/doi/full/10.1112/jlms.12507 requires payment or institutional subscription. A version is freely available on arXiv at https://arxiv.org/abs/2010.09200
The question is a special case of the problem which is labelled diophantine m-tuples in the literature, or more specifically you are asking for a diophantine quadruple with the property D(-1). Andrej Dujella has published numerous papers on the subject, and has a web page here on the subject, which puts the question into that context.
The question is still unsolved, but Dujella et. al. proves in their paper Effective solution of the D(-1)-quadruple conjecture that there can be only finitely many with $10^{10^{23}}$ being an upper bound on the maximum of $w, x, y , z$ in your notation.