Find all positive integers $x,y,z$ with $z$ odd, which satisfy the equation:

$$2018^x=100^y + 1918^z$$
We are given the equation:
\[ 2018^x = 100^y + 1918^z \]
where \( x, y, z \) are positive integers and \( z \) is odd. We need to find all such \( x, y, z \).

1. **Rewrite the equation using prime factorization:**
   \[ 2018 = 2 \times 1009 \]
   \[ 100 = 2^2 \times 5^2 \]
   \[ 1918 = 2 \times 959 \]
   Substituting these into the equation, we get:
   \[ (2 \times 1009)^x = (2^2 \times 5^2)^y + (2 \times 959)^z \]
   Simplifying further:
   \[ 2^x \times 1009^x = 2^{2y} \times 5^{2y} + 2^z \times 959^z \]

2. **Consider the equation modulo 3:**
   \[ 2018 \equiv 2 \pmod{3} \]
   \[ 100 \equiv 1 \pmod{3} \]
   \[ 1918 \equiv 2 \pmod{3} \]
   Thus:
   \[ 2^x \equiv 1^y + 2^z \pmod{3} \]
   \[ 2^x \equiv 1 + 2^z \pmod{3} \]
   Since \( z \) is odd, \( 2^z \equiv 2 \pmod{3} \):
   \[ 2^x \equiv 1 + 2 \equiv 0 \pmod{3} \]
   This implies \( x \) must be odd:
   \[ x \equiv 1 \pmod{2} \]

3. **Case 1: \( x > z \)**
   \[ 2^{x-z} \times 1009^x - 959^z = 2^{2y-z} \times 5^{2y} \]
   Since the right-hand side (RHS) is an integer, the left-hand side (LHS) must also be an integer. Given \( x \) is odd, \( 2y - z \geq 1 \) and \( x - z \geq 1 \). Considering modulo 2:
   \[ 959^z \equiv 1 \pmod{2} \]
   This leads to a contradiction since \( 959^z \equiv 1 \pmod{2} \) implies \( 959^z \) is odd, but \( 2^{x-z} \times 1009^x \) is even. Thus, \( x \leq z \).

4. **Case 2: \( x < z \)**
   \[ 1009^x - 2^{z-x} \times 959^z = 2^{2y-x} \times 5^{2y} \]
   Again, since the RHS is an integer, the LHS must also be an integer. Given \( x \) is odd, \( 2y - x \geq 1 \) and \( z - x \geq 1 \). Considering modulo 2:
   \[ 1009^x \equiv 1 \pmod{2} \]
   This leads to a contradiction since \( 1009^x \equiv 1 \pmod{2} \) implies \( 1009^x \) is odd, but \( 2^{z-x} \times 959^z \) is even. Thus, \( x \geq z \).

5. **Case 3: \( x = z \)**
   Substituting \( x = z \) into the original equation:
   \[ 2018^x = 100^y + 1918^x \]
   \[ 2^x \times 1009^x = 2^{2y} \times 5^{2y} + 2^x \times 959^x \]
   Dividing both sides by \( 2^x \):
   \[ 1009^x = 2^{2y-x} \times 5^{2y} + 959^x \]
   Assume \( 2y - x > 1 \). Considering modulo 4:
   \[ 1009^x \equiv 1 \pmod{4} \]
   \[ 959^x \equiv (-1)^x \pmod{4} \]
   Since \( x \) is odd:
   \[ 1009^x \equiv 1 \pmod{4} \]
   \[ 959^x \equiv -1 \pmod{4} \]
   This implies \( x \) is even, a contradiction. Thus, \( 2y - x = 1 \).

6. **Solve for \( x, y, z \):**
   \[ x = 2y - 1 = z \]
   Substituting \( x = z = 1 \):
   \[ 2018^1 = 100^1 + 1918^1 \]
   \[ 2018 = 100 + 1918 \]
   \[ 2018 = 2018 \]
   This is true, so \( x = y = z = 1 \) is a solution.

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