Determine all triples $(x,y,z)$ of integers greater than $1$ with the property that $x$ divides $yz-1$, $y$ divides $zx-1$ and $z$ divides $xy-1$.
1. Assume without loss of generality (WLOG) that \( x \ge y \ge z > 1 \).
2. Given the conditions:
   \[
   x \mid yz - 1, \quad y \mid zx - 1, \quad z \mid xy - 1
   \]
   we can write:
   \[
   \frac{yz-1}{x}, \quad \frac{zx-1}{y}, \quad \frac{xy-1}{z}
   \]
   are all integers. Therefore, their product is also an integer:
   \[
   \frac{(yz-1)(zx-1)(xy-1)}{xyz}
   \]
3. Expanding the product in the numerator:
   \[
   (yz-1)(zx-1)(xy-1) = yzx^2y^2z^2 - yzx^2y - yzx^2z - yzx^2 + yzx - yzx - yzx + 1
   \]
   Simplifying the expression:
   \[
   = x^2y^2z^2 - x^2yz - x^2yz - xy^2z - xyz^2 + xy + yz + zx - 1
   \]
4. Since \( xyz \) divides the above expression, we have:
   \[
   xyz \mid xy + yz + zx - 1
   \]
   This implies:
   \[
   xyz \le xy + yz + zx - 1 < xy + yz + zx \le 3xy
   \]
   Hence:
   \[
   z \le 3
   \]
5. Since \( z > 1 \), we have \( z = 2 \) or \( z = 3 \). Let's consider \( z = 2 \):
   \[
   2xy \mid xy + 2x + 2y - 1
   \]
   This implies:
   \[
   xy \le 2x + 2y \le 4y \implies xy \le 4x \implies y \in \{2, 3\}
   \]
6. If \( y = 2 \):
   \[
   2 = y \mid zx - 1 = 2x - 1
   \]
   This is a contradiction since \( 2x - 1 \) is odd and cannot be divisible by 2. Hence, \( y \neq 2 \).

7. If \( y = 3 \):
   \[
   x \mid yz - 1 = 6 - 1 = 5 \implies x = 5
   \]

Thus, the only solution for integers greater than 1 is \( (x, y, z) = (5, 3, 2) \).

The final answer is \( \boxed{ (5, 3, 2) } \)