2. Determine all positive integers $k$ for which there exist positive integers $r$ and $s$ that satisfy the equation
$$
\left(k^{2}-6 k+11\right)^{r-1}=(2 k-7)^{s}
$$
Solution: Clearly, if $r=1$, then $2 k-7=1$ or $2 k-7=-1$. Thus, two solutions are $k=4$ and $k=3$. Furthermore, notice that if $k=2$, then $3^{r-1}=(-3)^{s}$, which has a solution for $r$ and $s$. Thus, another solution is $k=2$.

For $r \geq 2$, notice that $k^{2}-6 k+11=(k-3)^{2}+2 \geq 2$ and $k^{2}-6 k+11>2 k-7$ because $(k-4)^{2}>-2$. Moreover, $k^{2}-6 k+11$ and $(2 k-7)$ have the same prime factors. Let $p$ be a prime factor of $k^{2}-6 k+11$ and $2 k-7$, then
$$
p \mid\left[\left(k^{2}-6 k+11\right)+(2 k-7)\right]=(k-2)^{2} \text {, which implies that } p \mid(k-2)
$$

Moreover, $p \mid[2(k-2)-(2 k-7)]=3$. Hence, $p=3$. This means that there are positive integers $m$ and $n$ with $m \geq n$ such that $k^{2}-6 k+11=3^{m}$ and $2 k-7=3^{n}$. Notice that
$$
4 \cdot 3^{m}=4(k-3)^{2}+8=(2 k-6)^{2}+8=\left(3^{n}+1\right)^{2}+8=3^{2 n}+2 \cdot 3^{n}+9 .
$$

Since $3^{n} \mid\left(3^{2 n}+2 \cdot 3^{n}+9\right)$, then $3^{n} \mid 9$ and $n \leq 2$. Hence, we only have two cases left.
- If $n=1$, then $2 k-7=3$ and $k=5$ and $3^{m}=6$, which is not possible.
- If $n=2$, then $2 k-7=9$ and $k=8$ and $3^{m}=27$, which means $m=3$.

Thus, $k=8$ is another solution. Therefore, $k=2,3,4$, and 8 are the solutions.