Find all $ x,y,z\in\mathbb{N}_{0}$ such that $ 7^{x} \plus{} 1 \equal{} 3^{y} \plus{} 5^{z}$.

[i]Alternative formulation:[/i] Solve the equation $ 1\plus{}7^{x}\equal{}3^{y}\plus{}5^{z}$ in nonnegative integers $ x$, $ y$, $ z$.
We need to find all nonnegative integer solutions \((x, y, z)\) to the equation:
\[ 7^x + 1 = 3^y + 5^z. \]

1. **Initial Observations:**
   - If \(x = 0\), then \(7^x + 1 = 1 + 1 = 2\). Thus, \(3^y + 5^z = 2\). The only solution in nonnegative integers is \(y = 0\) and \(z = 0\). Therefore, \((x, y, z) = (0, 0, 0)\) is a solution.
   - If \(x = 1\), then \(7^x + 1 = 7 + 1 = 8\). Thus, \(3^y + 5^z = 8\). The only solution in nonnegative integers is \(y = 1\) and \(z = 1\). Therefore, \((x, y, z) = (1, 1, 1)\) is a solution.

2. **Assume \(x, y, z > 0\) and \(y > 1\):**
   - Consider the equation modulo 3:
     \[ 7^x + 1 \equiv 3^y + 5^z \pmod{3}. \]
     Since \(7 \equiv 1 \pmod{3}\) and \(5 \equiv 2 \pmod{3}\), we have:
     \[ 1^x + 1 \equiv 0 \pmod{3} \Rightarrow 2 \equiv 2^z \pmod{3}. \]
     This implies \(z\) is odd.

3. **Consider the equation modulo 4:**
   \[ 7^x + 1 \equiv 3^y + 5^z \pmod{4}. \]
   Since \(7 \equiv -1 \pmod{4}\) and \(5 \equiv 1 \pmod{4}\), we have:
   \[ (-1)^x + 1 \equiv (-1)^y + 1 \pmod{4}. \]
   This implies \(x \equiv y \pmod{2}\).

4. **Consider the equation modulo 5:**
   \[ 7^x + 1 \equiv 3^y + 5^z \pmod{5}. \]
   Since \(7 \equiv 2 \pmod{5}\) and \(5 \equiv 0 \pmod{5}\), we have:
   \[ 2^x + 1 \equiv (-2)^y \pmod{5}. \]
   If \(x\) and \(y\) are even, let \(x = 2a\) and \(y = 2b\):
   \[ (-1)^a + 1 \equiv (-1)^b \pmod{5}. \]
   This is a contradiction since \(\{0, 2\} \cap \{-1, 1\} = \emptyset\). Hence, \(x\) and \(y\) are odd.

5. **Reading again \(2^x + 1 = 3^y\):**
   Since \((\{2, -2\} + 1) \cap \{3, -3\} = \{3\}\), we find \(x \equiv y \equiv 1 \pmod{4}\).

6. **Consider the equation modulo 16:**
   \[ 7^x + 1 \equiv 3^y + 5^z \pmod{16}. \]
   Since \(7^2 \equiv 1 \pmod{16}\) and \(3^4 \equiv 1 \pmod{16}\), we have:
   \[ 7^x \equiv 7 \pmod{16}, \quad 3^y \equiv 3 \pmod{16} \Rightarrow 5^z \equiv 5 \pmod{16} \Rightarrow z \equiv 1 \pmod{4}. \]
   Hence, \(x, y, z \equiv 1 \pmod{4}\).

7. **Combining \(3^y + 5^z \equiv 1 \pmod{7}\) with \(7^x + 1 \equiv 5^z \pmod{9}\):**
   - Consider the possible residues of \(x \mod 3\):
     - If \(x \equiv 0 \pmod{3}\), then \(7^x \equiv 1 \pmod{9} \Rightarrow 5^z \equiv 2 \pmod{9} \Rightarrow z \equiv 5 \pmod{6} \Rightarrow 5^z \equiv 3 \pmod{7} \Rightarrow 3^y \equiv 5 \pmod{7} \Rightarrow y \equiv 5 \pmod{6}\). Thus, \((x, y, z) \equiv (9, 5, 5) \pmod{12}\). But then \(7^9 + 1 \equiv 3^5 + 5^5 \pmod{13}\) is not true.
     - The other cases, \(x \equiv 1, 2 \pmod{3}\), are similar and left to the reader.

Thus, the only solutions are \((0, 0, 0)\) and \((1, 1, 1)\).

The final answer is \( \boxed{ (0, 0, 0) } \) and \((1, 1, 1)\).