We are tasked with finding all nonnegative integer solutions $(x, y, z, w)$ to the equation:
$$2^x \cdot 3^y - 5^z \cdot 7^w = 1.$$

First, we note that $x \geq 1$ because if $x = 0$, the left-hand side would be a fraction, which cannot equal 1.

### Case 1: $w = 0$
The equation simplifies to:
$$2^x \cdot 3^y = 1 + 5^z.$$

- **Subcase 1.1: $z = 0$**
$$2^x \cdot 3^y = 1.$$
This implies $x = 0$ and $y = 0$, but $x \geq 1$, so this is not possible.

- **Subcase 1.2: $z = 1$**
$$2^x \cdot 3^y = 6.$$
The possible solutions are $x = 1$ and $y = 1$.

- **Subcase 1.3: $z \geq 2$**
Taking modulo 4, we get:
$$2^x \cdot 3^y \equiv 1 \pmod{4}.$$
Since $2^x \equiv 0 \pmod{4}$ for $x \geq 2$, this leads to a contradiction.

### Case 2: $w = 1$
The equation simplifies to:
$$2^x \cdot 3^y = 1 + 7 \cdot 5^z.$$

- **Subcase 2.1: $z = 0$**
$$2^x \cdot 3^y = 8.$$
The possible solutions are $x = 3$ and $y = 0$.

- **Subcase 2.2: $z = 1$**
$$2^x \cdot 3^y = 36.$$
The possible solutions are $x = 2$ and $y = 2$.

- **Subcase 2.3: $z \geq 2$**
Taking modulo 8, we get:
$$2^x \equiv 1 \pmod{8}.$$
This implies $x = 0$, which is not possible.

### Case 3: $w \geq 2$
The equation becomes:
$$2^x \cdot 3^y = 1 + 5^z \cdot 7^w.$$

- **Subcase 3.1: $y = 0$**
$$2^x = 1 + 5^z \cdot 7^w.$$
Taking modulo 4, we get a contradiction for $x \geq 2$.

- **Subcase 3.2: $y \geq 1$**
Taking modulo 9, we get a contradiction for $z \geq 2$.

Thus, we have exhausted all possible cases and find the solutions to be:
$$\boxed{(1, 1, 1, 0), (2, 2, 1, 1), (1, 0, 0, 0), (3, 0, 0, 1)}.$$