positive integer ordered pair $(x,y,z)$ in factorial equations [1] Total no. of positive integer ordered pairs in $x!+y!+z! = x!\cdot y!$
[2] Total no. of positive integer ordered pairs in $x!+y!+z! = x!\cdot y!\cdot z!$
[3] If $x!\cdot y! = x!+y!+2^z$, Then no. of positive integer ordered pairs $(x,y,z)$ is
$\bf{My\; Try::}$ for $(1)$ one... Given $x!+y!+z! = x!\cdot y!$
Let $z = 1$, Then $\displaystyle x!+y! +1 = x!\cdot y!\Rightarrow y!\left(x!+1\right) = \left(x!+1\right)\Rightarrow y! = \frac{x!+1}{x!-1} = 1+\frac{2}{x!-1}$
So we get $x!-1 = 1\Rightarrow x=2$. So $\displaystyle y! = 3$. So no positive integer ordered pairs.
Similarly Let $z = 2$, Then $\displaystyle x!+y! +2 = x!\cdot y!\Rightarrow y!\left(x!-1\right) = x!+2\Rightarrow y! = \frac{x!+2}{x!-1} = 1+\frac{3}{x!-1}$
So we get $x!+1 = 3\Rightarrow x=2$. So $\displaystyle y! = 4$.So no positive integer ordered pairs.
Now i did not understand how can i bound value of $z$
Similar argument for $(2)$ and $(3)$
Help required
Thanks
 A: Solution to [1] follows.  Rearrange as
$$z!+1=x!y!-x!-y!+1=(x!-1)(y!-1) ~~~~(*)$$
Immediately, we see that $x,y\ge 2$. Since $(x!-1)$ has no factors in common with $x!$ (similarly for $y!, z!$), any solutions can't have "small" factors on either side.  Suppose that $x\ge y$.  Clearly $z\ge x$, so we  get $1\equiv (-1)(y!-1)\pmod{x!}$, which rearranges to $y!\equiv 0\pmod{x!}$.  Hence in fact $x=y$ and $z!+1$ is a square.  
This factorization does give the solution $z=4, x=y=3$., and the evidence that these are the only solutions is quite strong.  Compare this and this. A conjecture of Sierpinski or possibly Ramanujan, called Brocard's problem, is that only $z=4,5,7$ give square $z!+1$. Of these three candidates only $z=4$ is the square of a number of the form $x!-1$.  We will now solve the problem in question without resolving Brocard's problem, by using structure of the $(x!-1)$-type numbers.
Case 1: $x+1$ divides $x!$.  Then $(x+1)!$ divides $(x!)^2$.  We take (*) modulo $(x+1)!$ and get $1\equiv -x!-x!+1\pmod{(x+1)!}$, so $2(x!)\equiv 0$, a contradiction since $x\ge 2$. Note that this case is quite big, including $x+1=p^k$ for all $k\ge 3$, and also $x+1=uv$ for $u,v>1$ with $\gcd(u,v)=1$.
Case 2: $x+1=p$, a prime.  Now $x!\equiv -1\pmod{p}$, by Wilson's theorem.  We take (*) modulo $p$ and get $1\equiv (-2)^2\pmod{p}$.  Hence, $5\equiv 0\pmod{p}$, so $x=4$, which isn't prime.
Case 3: $x+1=p^2$, the square of some prime, yet $p^2$ does not divide $x!$.  Hence $2p>x$ so $p^2=x+1\le 2p$, which rearranges as $0\ge p^2-2p=p(p-2)$.  Only prime $p=2$ solves this, so $x+1=4$, which is the known solution above.
