To solve the system of equations in integers:
$$x^2 - y^2 = z$$
$$3xy + (x-y)z = z^2$$

we begin by analyzing the first equation. The expression $x^2 - y^2 = z$ can be factored as:
$$(x+y)(x-y) = z$$
Substituting this into the second equation, we have:
$$3xy + (x-y)z = z^2$$
Given the expression for $z$, we substitute $z = (x+y)(x-y)$ into the second equation:
$$3xy + (x-y)(x+y)(x-y) = ((x+y)(x-y))^2$$
Simplify and expand the terms:
$$3xy + (x^2-y^2)(x-y) = (x^2-y^2)^2$$

Now let's consider specific integer solutions by testing small values for simplification:

1. Consider $x = 2$, $y = 1$:
$$x^2 - y^2 = 2^2 - 1^2 = 4 - 1 = 3 \quad \Rightarrow \quad z = 3$$
$$3xy + (x-y)z = 3\cdot2\cdot1 + (2-1)\cdot3 = 6 + 3 = 9$$
$$z^2 = 3^2 = 9$$
Thus, $(2, 1, 3)$ is a solution.

2. Consider $x = 1$, $y = 2$:
$$x^2 - y^2 = 1^2 - 2^2 = 1 - 4 = -3 \quad \Rightarrow \quad z = -3$$
$$3xy + (x-y)z = 3\cdot1\cdot2 + (1-2)\cdot(-3) = 6 + 3 = 9$$
$$z^2 = (-3)^2 = 9$$
Thus, $(1, 2, -3)$ is a solution.

3. Consider $x = 1$, $y = 0$:
$$x^2 - y^2 = 1^2 - 0^2 = 1 - 0 = 1 \quad \Rightarrow \quad z = 1$$
$$3xy + (x-y)z = 3\cdot1\cdot0 + (1-0)\cdot1 = 0 + 1 = 1$$
$$z^2 = 1^2 = 1$$
Thus, $(1, 0, 1)$ is a solution.

4. Consider $x = 0$, $y = 1$:
$$x^2 - y^2 = 0^2 - 1^2 = 0 - 1 = -1 \quad \Rightarrow \quad z = -1$$
$$3xy + (x-y)z = 3\cdot0\cdot1 + (0-1)\cdot(-1) = 0 + 1 = 1$$
$$z^2 = (-1)^2 = 1$$
Thus, $(0, 1, -1)$ is a solution.

5. Consider $x = 0$, $y = 0$:
$$x^2 - y^2 = 0^2 - 0^2 = 0 - 0 = 0 \quad \Rightarrow \quad z = 0$$
$$3xy + (x-y)z = 3\cdot0\cdot0 + (0-0)\cdot0 = 0$$
$$z^2 = 0^2 = 0$$
Thus, $(0, 0, 0)$ is a solution.

The integer solutions to the system of equations are:
$$\boxed{\{(2, 1, 3), (1, 2, -3), (1, 0, 1), (0, 1, -1), (0, 0, 0)\}}$$