Find all triples $(x,y,z)$ of positive integers, with $z>1$, satisfying simultaneously that \[x\text{ divides }y+1,\quad y\text{ divides }z-1,\quad z\text{ divides }x^2+1.\]
We are given the following conditions for positive integers \(x, y, z\) with \(z > 1\):

1. \(x \text{ divides } y + 1\)
2. \(y \text{ divides } z - 1\)
3. \(z \text{ divides } x^2 + 1\)

We need to find all such triples \((x, y, z)\).

1. **Express \(y\) in terms of \(x\):**

   Since \(x \mid (y + 1)\), we can write:
   \[
   y + 1 = kx \quad \text{for some integer } k
   \]
   Therefore,
   \[
   y = kx - 1
   \]

2. **Express \(z\) in terms of \(y\):**

   Since \(y \mid (z - 1)\), we can write:
   \[
   z - 1 = my \quad \text{for some integer } m
   \]
   Therefore,
   \[
   z = my + 1
   \]
   Substituting \(y = kx - 1\) into the equation for \(z\), we get:
   \[
   z = m(kx - 1) + 1 = mkx - m + 1
   \]

3. **Substitute \(z\) into the third condition:**

   Since \(z \mid (x^2 + 1)\), we have:
   \[
   mkx - m + 1 \mid x^2 + 1
   \]
   Let \(d = mkx - m + 1\). We need \(d \mid (x^2 + 1)\). This implies:
   \[
   x^2 + 1 = c(mkx - m + 1) \quad \text{for some integer } c
   \]
   Expanding and rearranging, we get:
   \[
   x^2 + 1 = cmkx - cm + c
   \]
   \[
   x^2 - cmkx + cm + 1 - c = 0
   \]

4. **Analyze the quadratic equation:**

   The quadratic equation in \(x\) is:
   \[
   x^2 - cmkx + (cm + 1 - c) = 0
   \]
   For \(x\) to be an integer, the discriminant of this quadratic equation must be a perfect square. The discriminant \(\Delta\) is given by:
   \[
   \Delta = (cmk)^2 - 4(cm + 1 - c)
   \]
   \[
   \Delta = c^2m^2k^2 - 4cm - 4 + 4c
   \]
   \[
   \Delta = c^2m^2k^2 - 4cm + 4c - 4
   \]

5. **Check for perfect square solutions:**

   We need \(\Delta\) to be a perfect square. Let's test some small values for \(c, m, k\):

   - For \(c = 1\), \(m = 1\), \(k = 1\):
     \[
     \Delta = 1^2 \cdot 1^2 \cdot 1^2 - 4 \cdot 1 \cdot 1 + 4 \cdot 1 - 4 = 1 - 4 + 4 - 4 = -3 \quad (\text{not a perfect square})
     \]

   - For \(c = 1\), \(m = 1\), \(k = 2\):
     \[
     \Delta = 1^2 \cdot 1^2 \cdot 2^2 - 4 \cdot 1 \cdot 1 + 4 \cdot 1 - 4 = 4 - 4 + 4 - 4 = 0 \quad (\text{perfect square})
     \]
     This gives us a valid solution. Let's find \(x, y, z\):
     \[
     y = 2x - 1
     \]
     \[
     z = 1(2x - 1) + 1 = 2x - 1 + 1 = 2x
     \]
     Therefore, the triple \((x, y, z) = (x, 2x - 1, 2x)\) is a solution.

6. **Verify the solution:**

   - \(x \mid (2x - 1 + 1) \Rightarrow x \mid 2x \Rightarrow \text{True}\)
   - \(2x - 1 \mid (2x - 1) \Rightarrow \text{True}\)
   - \(2x \mid (x^2 + 1)\)

   For \(2x \mid (x^2 + 1)\), we need:
   \[
   x^2 + 1 = 2xk \quad \text{for some integer } k
   \]
   \[
   x^2 + 1 = 2xk
   \]
   \[
   x^2 - 2xk + 1 = 0
   \]
   The discriminant of this quadratic equation is:
   \[
   \Delta = (2k)^2 - 4 \cdot 1 \cdot 1 = 4k^2 - 4
   \]
   \[
   \Delta = 4(k^2 - 1)
   \]
   For \(\Delta\) to be a perfect square, \(k^2 - 1\) must be a perfect square. This implies \(k = 1\), giving us:
   \[
   x^2 - 2x + 1 = 0
   \]
   \[
   (x - 1)^2 = 0
   \]
   \[
   x = 1
   \]
   Therefore, the only solution is \(x = 1\), \(y = 1\), \(z = 2\).

The final answer is \( \boxed{ (1, 1, 2) } \).