Determine all triplet of non-negative integers $ (x,y,z) $ satisfy $$ 2^x3^y+1=7^z $$
To determine all triplets of non-negative integers \((x, y, z)\) that satisfy the equation
\[ 2^x 3^y + 1 = 7^z, \]
we will use Zsigmondy's Theorem and some elementary number theory.

1. **Check small values of \(z\):**
   - For \(z = 1\):
     \[ 7^1 = 7 \]
     \[ 2^x 3^y + 1 = 7 \]
     \[ 2^x 3^y = 6 \]
     The only solution in non-negative integers is \(x = 1\) and \(y = 1\). Thus, \((x, y, z) = (1, 1, 1)\) is a solution.

   - For \(z = 2\):
     \[ 7^2 = 49 \]
     \[ 2^x 3^y + 1 = 49 \]
     \[ 2^x 3^y = 48 \]
     Factorizing 48, we get:
     \[ 48 = 2^4 \cdot 3 \]
     Thus, \(x = 4\) and \(y = 1\). Therefore, \((x, y, z) = (4, 1, 2)\) is another solution.

2. **Use Zsigmondy's Theorem:**
   Zsigmondy's Theorem states that for any integers \(a > b > 0\), and \(n > 1\), there exists a prime \(p\) that divides \(a^n - b^n\) but does not divide \(a^k - b^k\) for any \(k < n\), with some exceptions. 

   Applying this to our problem:
   - Let \(a = 7\) and \(b = 1\), then \(7^z - 1\) must have a prime divisor \(p\) that does not divide \(7^k - 1\) for any \(k < z\).
   - For \(z \geq 3\), Zsigmondy's Theorem guarantees the existence of such a prime \(p\) that is not 2 or 3, which contradicts the form \(2^x 3^y\). Therefore, \(z\) cannot be 3 or greater.

3. **Conclusion:**
   Since \(z\) can only be 1 or 2, and we have already found the solutions for these cases, the only solutions are:
   \((x, y, z) = (1, 1, 1)\) and \((x, y, z) = (4, 1, 2)\).

The final answer is \( \boxed{ (1, 1, 1) } \) and \((4, 1, 2)\).