15. (25 points) Find all non-negative integer solutions to the equation
$$
x^{3}+y^{3}-x^{2} y^{2}-(x+y)^{2} z=0
$$
15. (1) When $x=0$, the equation becomes
$$
y^{3}-y^{2} z=0 \Rightarrow y^{2}(y-z)=0
$$
$\Rightarrow y=0$ or $y=z$.
Thus, $(0,0, m),(0, m, m)\left(m \in \mathbf{Z}_{+}\right)$ are solutions that satisfy the problem.
(2) When $y=0$, similarly,
$$
(0,0, m),(m, 0, m)\left(m \in \mathbf{Z}_{+}\right)
$$

are solutions that satisfy the problem.
(3) When $x, y \in \mathbf{Z}_{+}$, let $a=x+y, b=x y$.
Then the original equation becomes
$$
b^{2}+3 a b-a^{2}(a-z)=0 \text {. }
$$

Considering that $a, b, z$ are all integers, thus, the discriminant
$$
\Delta=9 a^{2}+4 a^{2}(a-z)=a^{2}(4 a+9-4 z)
$$

is a perfect square.
Since $4 a+9-4 z$ is odd, we can assume
$$
4 a+9-4 z=(2 t+1)^{2}(t \geqslant 0) \text {. }
$$

Then $a=t^{2}+t+z-2, b=\frac{-3 a+\sqrt{\Delta}}{2}=a(t-1)$.
Since $a, b \in \mathbf{Z}_{+}$, we know $t \geqslant 2$.
Next, consider the square number
$$
\begin{array}{l}
(x-y)^{2}=(x+y)^{2}-4 x y=a^{2}-4 b \\
=a^{2}-4 a(t-1) . \\
\text { Then }[a-2(t-1)-2]^{2} \leqslant a^{2}-4 a(t-1) \\
<[a-2(t-1)]^{2},
\end{array}
$$

and
$$
\begin{array}{l}
a^{2}-4 a(t-1) \neq[a-2(t-1)-1]^{2} . \\
\text { Hence } a^{2}-4 a(t-1)=[a-2(t-1)-2]^{2} \\
\Rightarrow a=t^{2} \Rightarrow t+z=2 .
\end{array}
$$

Therefore, $t=2, z=0$.
Thus, $a=b=4, x=y=2$.
At this point, the solution to the equation is $(2,2,0)$.
In summary, all non-negative integer solutions to the equation are
$$
(2,2,0),(0,0, m),(0, m, m),(m, 0, m) \text {, }
$$

where $m$ is any non-negative integer.
(Provided by Guo Min)