The Sequence $\{a_{n}\}_{n \geqslant 0}$  is defined by $a_{0}=1, a_{1}=-4$ and $a_{n+2}=-4a_{n+1}-7a_{n}$ , for $n \geqslant 0$. Find the number of positive integer divisors of $a^2_{50}-a_{49}a_{51}$.
1. Define the sequence $\{a_n\}_{n \geq 0}$ by the recurrence relation $a_{n+2} = -4a_{n+1} - 7a_{n}$ with initial conditions $a_0 = 1$ and $a_1 = -4$.
2. Consider the expression $\mathcal{B}_n = a_{n+1}^2 - a_n \cdot a_{n+2}$.
3. Substitute $a_{n+2} = -4a_{n+1} - 7a_{n}$ into $\mathcal{B}_n$:
   \[
   \mathcal{B}_n = a_{n+1}^2 - a_n \cdot (-4a_{n+1} - 7a_{n})
   \]
4. Simplify the expression:
   \[
   \mathcal{B}_n = a_{n+1}^2 + 4a_n a_{n+1} + 7a_n^2
   \]
5. Notice that $\mathcal{B}_n$ can be written as:
   \[
   \mathcal{B}_n = (a_{n+1} + 2a_n)^2 + 3a_n^2
   \]
6. From the recurrence relation, we can derive that $\mathcal{B}_n = 7 \mathcal{B}_{n-1}$.
7. Calculate the initial value $\mathcal{B}_0$:
   \[
   \mathcal{B}_0 = a_1^2 - a_0 \cdot a_2
   \]
   Given $a_0 = 1$, $a_1 = -4$, and $a_2 = -4a_1 - 7a_0 = 16 - 7 = 9$:
   \[
   \mathcal{B}_0 = (-4)^2 - 1 \cdot 9 = 16 - 9 = 7
   \]
8. Using the recurrence relation for $\mathcal{B}_n$, we have:
   \[
   \mathcal{B}_n = 7^n \mathcal{B}_0
   \]
   Therefore:
   \[
   \mathcal{B}_{49} = 7^{49} \cdot 7 = 7^{50}
   \]
9. The number of positive integer divisors of $7^{50}$ is given by the formula for the number of divisors of a number in the form $p^k$, which is $k+1$.
   \[
   \text{Number of positive integer divisors of } 7^{50} = 50 + 1 = 51
   \]

The final answer is $\boxed{51}$.