Solution: The given equation is \begin{equation}\label{eq:08Mar2025-1} \left( x^2 - 9x + 19 \right) ^{\left( x^2 - 15x + 56 \right) } = 1. \end{equation} We note that $x^y = 1$ holds in integers if and only if either of the following holds:

1. $x = 1$ and $y \in \mathbb{Z} $;
2. $x \in \mathbb{Z} \setminus \{ 0 \} $ and $y = 0$;
3. $x = - 1$ and $y$ is even.

We will take these three cases into consideration.

Case 1: Let \( x^2 - 9x + 19 = 1 \) and $x^2 - 15x + 56 \in \mathbb{Z} $. Thus, \begin{align*} x^2 - 9x + 19 = 1 & \implies x^2 - 9x + 18 = 0 \\ & \implies (x-3)(x-6) = 0 \\ & \implies x = 3 \text{ or } x = 6. \end{align*}

Case 2: Let \( x^2 - 9x + 19 \in \mathbb{Z} \setminus \{ 0 \} \) and $x^2 - 15x + 56 = 0$. Thus, \begin{align*} x^2 - 15x + 56 = 0 & \implies (x - 7) (x - 8) = 0 \\ & \implies x = 7 \text{ or } x = 8. \end{align*} Note that for $x = 7$ or $x = 8$, $x^2 - 9x + 19 \neq 0$.

Case 3: Let \( x^2 - 9x + 19 = -1 \) and \( x^2 - 15x + 56 \) is even. Then we have \begin{align*} x^2 - 9x + 19 = -1 & \implies x^2 - 9x + 20 = 0 \\ & \implies (x - 4)(x - 5) = 0 \\ & \implies x = 4 \text{ or } x = 5. \end{align*} Putting these values in $x^2 - 15x + 56 $, we get $12$ and $6$. Thus both values are possible. Therefore, the solution of the equation \eqref{eq:08Mar2025-1} in $\mathbb{Z} $ will be $3,4,5,6,7,8$.