Find all triples of natural numbers $ (x, y, z) $ that satisfy the condition $ (x + 1) ^ {y + 1} + 1 = (x + 2) ^ {z + 1}. $
1. **Rewrite the given equation:**
   \[
   (x + 1)^{y + 1} + 1 = (x + 2)^{z + 1}
   \]

2. **Apply Zsigmondy's theorem:**
   Zsigmondy's theorem states that for any integers \(a > b > 0\) and \(n > 1\), there exists a prime \(p\) (called a Zsigmondy prime) that divides \(a^n - b^n\) but does not divide \(a^k - b^k\) for any \(k < n\), with some exceptions. One of the exceptions is when \(a = 2\) and \(n = 3\), i.e., \(2^3 - 1 = 7\).

3. **Identify the exception case:**
   In our problem, we can set \(a = x + 1\) and \(b = -1\). The exception case for Zsigmondy's theorem is when \((x + 1)^{y + 1} = 2^3 = 8\).

4. **Solve for \(x\) and \(y\):**
   \[
   (x + 1)^{y + 1} = 8 \implies (x + 1)^{y + 1} = 2^3
   \]
   This implies \(x + 1 = 2\) and \(y + 1 = 3\). Therefore:
   \[
   x + 1 = 2 \implies x = 1
   \]
   \[
   y + 1 = 3 \implies y = 2
   \]

5. **Substitute \(x = 1\) and \(y = 2\) into the original equation:**
   \[
   (1 + 1)^{2 + 1} + 1 = (1 + 2)^{z + 1}
   \]
   Simplify:
   \[
   2^3 + 1 = 3^{z + 1}
   \]
   \[
   8 + 1 = 3^{z + 1}
   \]
   \[
   9 = 3^{z + 1}
   \]
   \[
   3^2 = 3^{z + 1}
   \]
   Therefore:
   \[
   z + 1 = 2 \implies z = 1
   \]

6. **Verify the solution:**
   The solution \((x, y, z) = (1, 2, 1)\) satisfies the original equation:
   \[
   (1 + 1)^{2 + 1} + 1 = (1 + 2)^{1 + 1}
   \]
   \[
   2^3 + 1 = 3^2
   \]
   \[
   8 + 1 = 9
   \]
   \[
   9 = 9
   \]
   This confirms that the solution is correct.

\(\blacksquare\)

The final answer is \( \boxed{ (x, y, z) = (1, 2, 1) } \)