2. Solve the equation

$$
\left(n^{4}-9 n^{2}+19\right)^{n^{3}+14 n^{2}+33 n}=1
$$

in the set of integers.
2. For the expression on the left to be equal to 1 for integer bases and integer exponents, one of the following conditions must hold: the exponent is 0 and the base is non-zero; the base is 1 and the exponent is arbitrary; the base is -1 and the exponent is an even number. Let's now consider these possibilities separately.

- $n^{3}+14 n^{2}+33 n=0$ and $n^{4}-9 n^{2}+19 \neq 0$: We have $0=n^{3}+14 n^{2}+33 n=n\left(n^{2}+14 n+33\right)$, so, in addition to $n=0$, the solutions are also $n=\frac{-14 \pm \sqrt{14^{2}-4 \cdot 33}}{2}=\frac{-14 \pm 8}{2}$, i.e., $n=-3$ and $n=-11$. It is directly verified that all these solutions indeed satisfy the condition $n^{4}-9 n^{2}+19 \neq 0$.
- $n^{4}-9 n^{2}+19=1$: The equation $n^{4}-9 n^{2}+19=1$ can be reduced to $t^{2}-9 t+18=0$ by the substitution $n^{2}=t$, whose solutions are $t=\frac{9 \pm \sqrt{9^{2}-4 \cdot 18}}{2}=\frac{9 \pm 3}{2}$, i.e., $t=6$ and $t=3$. Considering $n= \pm \sqrt{t}$, we do not obtain integer solutions for $n$ here.
- $n^{4}-9 n^{2}+19=-1$ and $n^{3}+14 n^{2}+33 n$ is an even number: The equation $n^{4}-9 n^{2}+19=-1$ can be reduced to $t^{2}-9 t+20=0$ by the substitution $n^{2}=t$, whose solutions are $t=\frac{9 \pm \sqrt{9^{2}-4 \cdot 20}}{2}=\frac{9 \pm 1}{2}$, i.e., $t=5$ and $t=4$. In the first case, we do not obtain integer solutions for $n$, while in the second case, we get $n=2$ and $n=-2$. We observe that for both these values, the expression $n^{3}+14 n^{2}+33 n$ is an even number, so these are solutions to the equation in the problem statement.

Therefore, the solutions to the original equation are: $n \in\{-11,-3,-2,0,2\}$.