We are tasked with finding all non-negative integer solutions $(x, y, z)$ to the equation:

$$2^x + 9 \cdot 7^y = z^3$$

Given that our solution must satisfy integer constraints and each variable is greater than or equal to zero, we will systematically explore potential solutions.

### Step 1: Analyze small values of $z$
Let's begin by examining small values for $z$ since the cubic function grows rapidly and only a finite number of candidates will exist.

#### Sub-step: Calculate for $z = 0, 1, 2, 3, 4$
$$z = 0: \quad 2^x + 9 \cdot 7^y = 0^3 = 0$$

This forces $2^x = 0$ and $9 \cdot 7^y = 0$, which is impossible for $x, y \geq 0$.

$$z = 1: \quad 2^x + 9 \cdot 7^y = 1^3 = 1$$

This can only be satisfied by $2^x = 1$ and $9 \cdot 7^y = 0$, which is impossible for integer $y \geq 0$.

$$z = 2: \quad 2^x + 9 \cdot 7^y = 2^3 = 8$$

Testing for $x = 0, 1, 2, 3$:
- $x = 3$: $2^3 = 8$, so $9 \cdot 7^y = 0$, $y$ cannot be satisfied for $y \geq 0$.

$$z = 3: \quad 2^x + 9 \cdot 7^y = 3^3 = 27$$

Testing for small values of $x$ and $y$:
- $x = 0$: $2^0 + 9 \cdot 7^0 = 1 + 9 = 10$, not $27$.
- $x = 1$: $2^1 + 9 \cdot 7^0 = 2 + 9 = 11$, not $27$.
- $x = 2$: $2^2 + 9 \cdot 7^0 = 4 + 9 = 13$, not $27$.

Trying with different values of $y$, we find that no combination of $(x, y)$ results in $27 - 2^x = 9 \cdot 7^y$.

$$z = 4: \quad 2^x + 9 \cdot 7^y = 4^3 = 64$$

For $x = 0$, solve $1 + 9 \cdot 7^y = 64$:
- $9 \cdot 7^y = 63$ which yields $7^y = 7$ and hence $y = 1$.

The solution $(x, y, z) = (0, 1, 4)$ satisfies $2^0 + 9 \cdot 7^1 = 1 + 63 = 64 = 4^3$.

### Conclusion
Therefore, the only solution with $x, y, z \geq 0$ satisfying the equation is:

$$\boxed{(0, 1, 4)}$$