Determine all tuples of positive integers $(x, y, z, t)$ such that: 
$$ xyz = t!$$ 
$$ (x+1)(y+1)(z+1) = (t+1)!$$ 
holds simultaneously.
To solve the problem, we need to find all tuples of positive integers \((x, y, z, t)\) such that the following two equations hold simultaneously:
\[ xyz = t! \]
\[ (x+1)(y+1)(z+1) = (t+1)! \]

We will use the function \( f(x) = 1 + \frac{1}{x} \) and analyze the possible values of \(z\) and \(t\).

1. **Case \(z \geq 4\):**
   - If \(z \geq 4\), then \(f(z) \leq 1.25\).
   - Therefore, \(f(x)f(y)f(z) \leq 1.25^3 = 1.953125 < 2\).
   - Since \(f(x)f(y)f(z) > 1\), there are no solutions in this case.

2. **Case \(z = 3\):**
   - Here, \(f(z) = \frac{4}{3}\).
   - Thus, \(f(x)f(y)f(z) \leq \left(\frac{4}{3}\right)^3 = \frac{64}{27} \approx 2.37 < 3\).
   - We must have \(t = 1\) because \((t+1)! = 2\).
   - Therefore, \(f(x)f(y) = \frac{2}{f(3)} = \frac{2}{\frac{4}{3}} = \frac{3}{2}\).
   - Solving for \(x\) and \(y\) such that \(f(x)f(y) = \frac{3}{2}\), we find:
     - \(f(x) = \frac{3}{2}\) and \(f(y) = 1\) or vice versa.
     - This gives \((x, y) = (8, 3)\) or \((5, 4)\).

3. **Case \(z = 2\):**
   - Here, \(f(z) = \frac{3}{2}\).
   - Thus, \(f(x)f(y)f(z) \leq \left(\frac{3}{2}\right)^3 = \frac{27}{8} = 3.375 < 4\).
   - We must have \(t = 2\) because \((t+1)! = 6\).
   - Therefore, \(f(x)f(y) = \frac{6}{f(2)} = \frac{6}{\frac{3}{2}} = 4\).
   - Solving for \(x\) and \(y\) such that \(f(x)f(y) = 4\), we find:
     - \(f(x) = 2\) and \(f(y) = 2\) or vice versa.
     - This gives \((x, y) = (3, 2)\), \((7, 6)\), \((9, 5)\), or \((15, 4)\).

4. **Case \(z = 1\):**
   - Here, \(f(z) = 2\).
   - Thus, \(f(x)f(y)f(z) \leq 2 \cdot f(x)f(y) \leq 8\).
   - We must have \(t = 3\) because \((t+1)! = 24\).
   - Therefore, \(f(x)f(y) = \frac{24}{f(1)} = \frac{24}{2} = 12\).
   - Solving for \(x\) and \(y\) such that \(f(x)f(y) = 12\), we find:
     - \(f(x) = 4\) and \(f(y) = 3\) or vice versa.
     - This gives \((x, y) = (1, 1)\), \((2, 1)\), \((3, 2)\), \((8, 3)\), or \((5, 4)\).

Thus, the tuples are any permutation of the first 3 entries of:
\[
(8, 3, 3, 1), (5, 4, 3, 1), (3, 2, 2, 2), (7, 6, 2, 1), (9, 5, 2, 1), (15, 4, 2, 1), (1, 1, 1, 7), (2, 1, 1, 5), (3, 2, 1, 3), (8, 3, 1, 2), (5, 4, 1, 2).
\]

The final answer is \( \boxed{ (8, 3, 3, 1), (5, 4, 3, 1), (3, 2, 2, 2), (7, 6, 2, 1), (9, 5, 2, 1), (15, 4, 2, 1), (1, 1, 1, 7), (2, 1, 1, 5), (3, 2, 1, 3), (8, 3, 1, 2), (5, 4, 1, 2) } \).