This question was also N1 JBMO 2022 TST test of Azerbaijan, For the sketch of my proof, Prove that $a,b,c$ are odd, use $\mod 4$ and do casework.

WLOG $a\le b\le c$ and suppose $ab\ge 5$. Then each expression is divisible by $3$, so we have $(abc)^2\equiv (ab)\cdot (bc)\cdot (ca)\equiv (-1)^3\equiv 2\pmod{3}$, contradiction, as $2$ isn't a QR$\pmod{3}$. Thus, we must have $ab=1$ or $ab=5$. If $a=b=1$, then we equivalently need $c=n!-1$ for some integer $n\ge 2$. If $a=1, b=5$, then note that because $5\nmid 5c+1$ and $5c+1\ge 5\cdot 5+1=26$, $5c+1$, $5c+1$ cannot be a factorial. Thus, the only solutions for $(a,b,c)$ are permutations of $(1,1,n!-1)$ for any integer $n\ge 2$.

Weirdly, this was in the Kosovo TST Suppose, by way of contradiction, that $ab+1=k!, bc+1=t!, ca+1=s!$ and that $s,k,t\geq 3.$ Then we have that $3\mid t!,s!,k!$ therefore $ab\equiv bc\equiv ca\equiv -1\pmod 3.$ Hence we get $a^2b^2c^2\equiv -1\pmod 3$ however this is impossible thus we must have, WLOG, $t<3.$ $t=0,1$ can't be since $a,b,c\in\mathbb{N}.$ Say $t=2$ then we have $bc+1=2!\Rightarrow b=c=1.$ Then $a=k!-1$ for any $k>1.$ So the solutions are of the form $\{(1,1,k!-1),(1,k!-1,1),(k!-1,1,1)\}$ for $k\geq2.$

ab+1=x!,bc+1=y!,ac+1=z!. suppose that x,y,z>=3; we get that ab=2 (mod 3),(1)WLOG a = 2 (mod 3) , c = 1 (mod 3) (2)bc = 2 (mod 3) and because of (1) => b = 2 (mod 3) but ab = 2 (mod 3) which is impossible beacuse a = b = 2 (mod 3) so WLOG x<3 <=> ab+1 = 2 <=> a=b=1 => c=y!-1=z!-1; so solutions are permutations of (1,1,y!-1)

Let all of them $>1$ $\Rightarrow ab+1,bc+1,ac+1 \geq 5 $ $ab+1,bc+1,ac+1 \geq 6 $$\Rightarrow$ $ab+1,bc+1,ac+1=0 \mod {3}$ $\Rightarrow a,b,c \ne 0 \mod{3}$ $a,b,c \ne 1 \mod{3}$ let $a =2 \mod {3}$ $\Rightarrow$$b,c = 1\mod {3}$ so its impossible $\Rightarrow $ one of $a,b,c$ need $=1$ let $a=1$ $b+1,c+1,bc+1$ let $b,c>1$ $b=k!-1,c=l!-1$,$bc+1=l!+k!-1=p!$ But we can see $l!,k!,p! =0 \mod{2}$ but $l!+k!-1 \ne 0\mod{2}$ so its impossible so some of $b,c=1$ $1)$ $b=1$$\Rightarrow$ $c=p!-1$ or $b,c=1$

Let $ab+1=r!, bc+1=s!$, and $ac+1=t!$. WLOG, let $c\geq{b}\geq{a}$. Then, $s!\equiv t!\equiv 0 \pmod{r!}$ and $ab\equiv bc\equiv ca\equiv -1\pmod{r!} \Longrightarrow a(c-b)\equiv b(c-a)\equiv c(b-a)\equiv 0 \pmod{r!}$. However, since $a,b\mid{r!-1}$, $\gcd(a, r!)=\gcd(b, r!)=1$, $(c-b)\equiv (c-a)\equiv (b-a)\equiv 0\pmod{r!} \Longrightarrow a\equiv b\equiv c\equiv x\pmod{r!}$. $ab+1\equiv x^2+1\equiv 0\pmod{r!}$ If $r\geq{3}, x^2\equiv 2\pmod{3}$, which is not possible. Therefore, $r=2$, which means $a=b=1$. Plugging this into our initial equations, the other two equations become $c+1=s!$.Therefore, $(a,b,c)=(1,1,s!-1)$ and permutations where $s>1$.

We claim that the only solutions are $\boxed{(k!-1,1,1)}$ (and its permutations), where $k\in\mathbb{N}_{>1}$. WLOG assume $a\geq b\geq c$. Then, we have $$ab+1\geq ac+1\geq bc+1.$$We know that if $m!\leq k!$then $m!\mid k!$. So, $$bc+1\mid ab+1\Longrightarrow bc+1\mid b(a-c)$$and $$bc+1\mid ac+1\Longrightarrow bc+1\mid c(a-b).$$Note that $\gcd(bc+1,c)\leq\gcd(bc,bc+1)=1$. So, $\gcd(bc+1,c)=\gcd(bc+1,b)=1$. We conclude that $bc+1\mid a-c$ and $bc+1\mid a-b\Longrightarrow bc+1\mid b-c\Longrightarrow b=c\text{ or }b-c\geq bc+1$. The latter is $(b+1)(c-1)\leq-1$, which gives no solutions. So, $b=c\Longrightarrow b^2+1=m!\Longrightarrow m>1$. If $m\geq3$ then $3\mid m!\Longrightarrow b^2\equiv2\pmod{3}$, impossible. So, $m=2$ and $b=c=1$. Note that $ab+1=a+1=k!$. So, $a=k!-1$. So, the only triples $(a,b,c)$ are the aforementioned.

let's assume that ab+1=m! bc+1=l! ca+1=p! and let {l, p, m}> or = 4 since that is the case we get ab≡bc≡ca≡3 we get that a^2b^2c^2≡-1≡3 mod 4 and that is not possible hence we get a contradiction => our assumption wasn't correct hence let's assume that l<4 => if l=1 this isn't possible since ab+1=1 => ab=0 this isn't possible since a and b are positive integers II if l=2 we get that ab=1 => a=1=b => c=n!-1 this is a valid solution(note this works for all permutations of a,b and c) III if l=3 we get that BGO a=1 and b=5 (if we exclude the n!-1 solution) we get that m and p are > or = to 3 hence we get that 2c≡c≡1 mod 3 =>2≡1 mod 3 contradiction so the only solutions are (1, 1, n!-1), (1, n!-1, 1), (n!-1, 1, 1) for all positive integers(note n!-1/ n has to be bigger or equal to 2).

WLOG $a \le b \le c$, and we assume that $ab \ge 5$. So each expression is divisible by $3$ so $(abc)^2 \equiv ab\cdot bc\cdot ca \equiv 2\pmod{3}$ so contradiction so $ab=1$ or $ab=5.$ Case 1: If $ab=1$, $a=b=1$, so $c=k!-1$, where $k \ge 2.$ Case 2: If $ab=5$: $a=1$ and $b=5$, so $5c+1$ and $c+1$ need to be factorials. Note that $\frac{5c+1}{c+1}$ gets very close to $5$ as $c$ gets bigger, testing $5, 23, 119$ none of them work, and now the quotient of 2 consecitive factorials is at least $6$, but the fraction is less than $6$, so if $ab=5$ we have no solutions. So the only solution is $\boxed{(1,1, k!-1)} k\ge 2$ and permutations.

Solved with DeliriousMink1 WLOG $a\geq b\geq c$ if $ab\geq 5$ then \begin{align*} ab\equiv -1 &\pmod 3 \\ bc\equiv -1 &\pmod 3 \\ ca\equiv -1 &\pmod 3 \\ \end{align*}Therefore $(abc)^2\equiv -1 \equiv 2\pmod 3$ so $ab\leq 5$, so $ab=1$ or $ab=5$, therefore we have two solutions $\boxed{a=1, b=1 \hspace{0.1cm} \text{and} \hspace{0.1cm} c=n!-1}$, the second case gives $a=1, b=5$ and $5c+1$ a factorial but is impposible for $c\geq 5$ as $5\mid 5c+1$ so no solutions in this case

The answer is all triples $(1,1,k!-1)$ for some positive integer $k$ (and cyclic variations). It is easy to see that these triples indeed satisfy the given equation. We now show that they are the only ones. WLOG, assume that $a \ge b \ge c$. It is then not hard to see that $ab+1 \ge ac+1 \ge bc+1$ so since all these numbers are factorials, we have $ca=1 \mid bc+1 \mid ab+1$. But then, \begin{align*} bc+1 & \mid ab +1 \\ bc+1 & \mid b(a-c)\\ bc +1 & \mid a-c \end{align*}since $\gcd(bc+1,b)=1$. Thus, $a=c$ or $a \ge bc+c+1$. Similarly, using $ac+1 \mid ab+1$ we can show that $b=c$ or $b \ge ac+c+1$. Now, if $b \neq c$, then clearly $a\neq c$ as well so in this case we have, \[bc+c+1\le a\le \frac{b-c-1}{c}\]which then rewrites to, \[b-c-1 \ge bc^2 + c^2 +c \ge b + c +1 > b - c -1\]which is a clear contradiction. Thus, we must have $b=c$. But then, $b^2+1=n!$ for some positive integer $n$ which implies $n<3$ (since $3\nmid b^2+1$ for all positive integers $b$). So, $n=1$ (which has no solutions) or $n=2$ which yields $b=c=1$. Then, it is immediate that $a=k!-1$ for some positive integer $k$ which finishes the proof.

We claim that only $\boxed{(a, b, c) = (1, 1, k!-1) \text{ and cyclic permutations }}$ work ($k>1$). Note that if WLOG $a = b$, we have $3 \nmid a^2 + 1 = x! \implies a = 1$, so $c = k!-1$ for some $k$. Else WLOG $a < b < c$. Then we have $ab+1 \mid bc+1 \implies ab+1\mid c-a$. Similarly, $ab+1 \mid c-b$. So $ab + 1 \mid \gcd(c-a, c-b) = \gcd(c-b, b-a) \mid b-1$. In particular, $ab + 1 \le b-a$, which isnt possible. $\square$