For each positive integer $k$ find the number of solutions in nonnegative integers $x,y,z$ with $x\le y \le z$ of the equation
$$8^k=x^3+y^3+z^3-3xyz$$
To find the number of solutions in nonnegative integers \(x, y, z\) with \(x \le y \le z\) for the equation 
\[ 8^k = x^3 + y^3 + z^3 - 3xyz, \]
we will use mathematical induction.

1. **Base Case: \(k = 0\)**
   \[
   8^0 = 1 = x^3 + y^3 + z^3 - 3xyz.
   \]
   We need to find nonnegative integer solutions \((x, y, z)\) such that \(x \le y \le z\). 

   Consider the equation:
   \[
   x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - xz - yz).
   \]
   For \(1 = (x + y + z)(x^2 + y^2 + z^2 - xy - xz - yz)\), we need to find the values of \(x, y, z\) that satisfy this equation. 

   The only solution in nonnegative integers is:
   \[
   (x, y, z) = (0, 0, 1).
   \]
   Therefore, for \(k = 0\), there is exactly 1 solution.

2. **Inductive Step:**
   Assume that for \(k = n-1\), there are \(n\) solutions. We need to show that for \(k = n\), there are \(n+1\) solutions.

   Given the equation for \(k = n-1\):
   \[
   8^{n-1} = x^3 + y^3 + z^3 - 3xyz,
   \]
   we assume there are \(n\) solutions \((x_i, y_i, z_i)\) for \(i = 1, 2, \ldots, n\).

   Now consider the equation for \(k = n\):
   \[
   8^n = x^3 + y^3 + z^3 - 3xyz.
   \]
   We can generate solutions for \(k = n\) by multiplying each solution \((x_i, y_i, z_i)\) of \(k = n-1\) by 2:
   \[
   (2x_i, 2y_i, 2z_i).
   \]
   This gives us \(n\) solutions for \(k = n\).

   Additionally, for any solution \((x, y, z)\) where \(x, y, z\) are all even, dividing each by 2 gives a solution for \(k = n-1\). Since we have \(n\) solutions for \(k = n-1\), we get \(n\) even-even-even solutions for \(k = n\).

   Therefore, we have \(n + 1\) solutions for \(k = n\).

By induction, we have shown that for any positive integer \(k\), the number of solutions in nonnegative integers \(x, y, z\) with \(x \le y \le z\) of the equation \(8^k = x^3 + y^3 + z^3 - 3xyz\) is \(k + 1\).

\(\blacksquare\)

The final answer is \( \boxed{ k + 1 } \).