Example 3. Solve the system of equations in integers:

$$
\left\{\begin{array}{l}
x^{2}-y^{2}-z^{2}=1 \\
y+z-x=3
\end{array}\right.
$$
Solution. Express $x$ in terms of $y$ and $z$ from the second equation and substitute into the first:

$$
y z-3 y-3 z+4=0
$$

From this,

$$
y=\frac{3 z-4}{z-3}=3+\frac{5}{z-3}
$$

Since $y \in \mathbf{Z}$, then $\frac{5}{z-3} \in \mathbf{Z}$. This is only possible if $z-3= \pm 1$ and $z-3= \pm 5$.

Thus, we get: $z_{1}=4, z_{2}=2, z_{3}=8, z_{4}=-2$, then find the corresponding values of $x$ and $y$. Therefore, the solution to the system can be written as

$$
\{(9 ; 8 ; 4),(-3 ;-2 ; 2),(9 ; 4 ; 8),(-3 ; 2 ;-2)\}
$$