How many integer triples $(x,y,z)$ are there such that \[\begin{array}{rcl} x - yz^2&\equiv & 1 \pmod {13} \\ xz+y&\equiv& 4 \pmod {13}  \end{array}\] where $0\leq x < 13$, $0\leq y <13$, and $0\leq z< 13$?

$ 
\textbf{(A)}\ 10
\qquad\textbf{(B)}\ 23
\qquad\textbf{(C)}\ 36
\qquad\textbf{(D)}\ 49
\qquad\textbf{(E)}\ \text{None of above}
$
To solve the problem, we need to find the number of integer triples \((x, y, z)\) that satisfy the following system of congruences modulo 13:
\[
\begin{array}{rcl}
x - yz^2 &\equiv & 1 \pmod{13} \\
xz + y &\equiv & 4 \pmod{13}
\end{array}
\]
where \(0 \leq x < 13\), \(0 \leq y < 13\), and \(0 \leq z < 13\).

1. **Express \(x\) in terms of \(y\) and \(z\):**
   From the first congruence, we have:
   \[
   x \equiv 1 + yz^2 \pmod{13}
   \]

2. **Substitute \(x\) into the second congruence:**
   Substitute \(x\) from the first congruence into the second congruence:
   \[
   (1 + yz^2)z + y \equiv 4 \pmod{13}
   \]
   Simplify the equation:
   \[
   z + yz^3 + y \equiv 4 \pmod{13}
   \]
   Combine like terms:
   \[
   y(z^3 + 1) + z \equiv 4 \pmod{13}
   \]
   Rearrange to solve for \(y\):
   \[
   y(z^3 + 1) \equiv 4 - z \pmod{13}
   \]

3. **Analyze the equation \(y(z^3 + 1) \equiv 4 - z \pmod{13}\):**
   We need to determine when \(z^3 + 1 \equiv 0 \pmod{13}\). This occurs when \(z^3 \equiv -1 \pmod{13}\).

4. **Find values of \(z\) such that \(z^3 \equiv -1 \pmod{13}\):**
   We need to check which values of \(z\) satisfy this condition. We can test each value of \(z\) from 0 to 12:
   \[
   \begin{array}{c|c}
   z & z^3 \pmod{13} \\
   \hline
   0 & 0 \\
   1 & 1 \\
   2 & 8 \\
   3 & 27 \equiv 1 \\
   4 & 64 \equiv 12 \\
   5 & 125 \equiv 8 \\
   6 & 216 \equiv 8 \\
   7 & 343 \equiv 5 \\
   8 & 512 \equiv 5 \\
   9 & 729 \equiv 1 \\
   10 & 1000 \equiv 12 \\
   11 & 1331 \equiv 12 \\
   12 & 1728 \equiv 12 \\
   \end{array}
   \]
   We see that \(z^3 \equiv -1 \pmod{13}\) when \(z \equiv 4, 10, 12 \pmod{13}\).

5. **Case analysis for \(z\):**
   - **Case \(z \equiv 4\):**
     \[
     y(4^3 + 1) \equiv 4 - 4 \pmod{13}
     \]
     \[
     y(64 + 1) \equiv 0 \pmod{13}
     \]
     \[
     y \cdot 65 \equiv 0 \pmod{13}
     \]
     Since \(65 \equiv 0 \pmod{13}\), any value of \(y\) works. There are 13 possible values for \(y\) (0 to 12). For each \(y\), \(x\) is uniquely determined by \(x \equiv 1 + yz^2 \pmod{13}\). Thus, there are 13 solutions in this case.

   - **Case \(z \equiv 10\) or \(z \equiv 12\):**
     \[
     y(z^3 + 1) \equiv 4 - z \pmod{13}
     \]
     For \(z = 10\):
     \[
     y(10^3 + 1) \equiv 4 - 10 \pmod{13}
     \]
     \[
     y(1000 + 1) \equiv -6 \pmod{13}
     \]
     \[
     y \cdot 1001 \equiv -6 \pmod{13}
     \]
     Since \(1001 \equiv 0 \pmod{13}\), there is no solution because \(-6 \not\equiv 0 \pmod{13}\).

     For \(z = 12\):
     \[
     y(12^3 + 1) \equiv 4 - 12 \pmod{13}
     \]
     \[
     y(1728 + 1) \equiv -8 \pmod{13}
     \]
     \[
     y \cdot 1729 \equiv -8 \pmod{13}
     \]
     Since \(1729 \equiv 0 \pmod{13}\), there is no solution because \(-8 \not\equiv 0 \pmod{13}\).

6. **Count the total number of solutions:**
   - There are 13 solutions when \(z \equiv 4\).
   - There are no solutions when \(z \equiv 10\) or \(z \equiv 12\).

   Therefore, the total number of solutions is:
   \[
   13 + 0 + 0 = 13
   \]

The final answer is \(\boxed{13}\)