A positive integer $k$ is called strong if there exists two positive integers $r$ and $s$ satisfying the Diophantine equation $(1) \;\; (k^2 - 6k + 11)^{r-1} = (2k - 7)^s$. If $k=1$, then $6^{r-1} = (-5)^s$ by equation (1), yielding $5 | 6$. Hence $k=1$ is not strong by contradiction. If $k=2$, then $3^{r-1} = (-3)^s$, which is satisfied when $(r,s)=(3,2)$. Therefore $k=2$ is strong. Assume $r=1$. Then $(2k - 7)^s = 1$ by equation (1), yielding $|2k - 7|=1$. Hence $k \in \{3,4\}$ and $s=2$ satisfies equation (1). Thus $k=3$ and $k=4$ are both strong. Next assume $r \geq 2$ and $k \geq 5$. Hence $2k - 7 \geq 2 \cdot 5 - 7 = 10 - 7 = 3$, implying there exists an odd prime divisor $p$ of $2k-7$. Consequently by equation (1) we obtain $p \mid k^2 - 6k + 11$, which means (since $p$ is odd and $2k \equiv 7 \pmod{p}$) $0 \equiv 4(k^2 - 6k + 11) = (2k)^2 - 12 \cdot (2k) + 44 \equiv 7^2 - 12 \cdot 7 + 44 = 49 - 84 + 44 = 9 \pmod{p}$, yielding $p \mid 3^2$. Therefore $p=3$, which according to equation (1) yields there are two positive integers $u$ and $v$ s.t. $(2) \;\; (k - 3)^2 + 2 = 3^u$, $(3) \;\; 2k - 7 = 3^v$. Hence by equation (2) $(2k)^2 - 12 \cdot (2k) + 44 = 4 \cdot 3^u$, which combined with equation (3) give us $(3^v + 7)^2 - 12 \cdot (3^v + 7) + 44 = 4 \cdot 3^u$, i.e. $(4) \;\; 3^{2v} + 2 \cdot 3^v + 9 = 4 \cdot 3^u$, or alternatively $(5) \;\; 3^{2v-2} + 2 \cdot 3^{v-2} + 1 = 4 \cdot 3^{u-2}$. Seeing that $k \geq 5$, we obtain by equation (2) $3^u = (k - 3)^2 + 2 \geq (5 - 3)^2 + 2 = 2^2 + 2 = 4 + 2 = 6 > 3$, which give us $u \geq 2$. Hence, since $v \geq 1$, we have $3^{v-2} \in \mathbb{N}$ by equation (5), i.e. $v \geq 2$. If $u,v \geq 3$, then $3 | 1$ by equation (5), implying $u=2$ or $v=2$. If $u=2$, then according to equation (4) we obtain $(3^v + 1)^2 = 28$, implying $u \neq 2$. Hence $v=2$, yielding $2k-7=3^v=3^2=9$, i.e. $k=8$. Inserting $v=2$ in equation(5), the result is $4 \cdot 3^{u-2} = 12$, i.e. $3^{u-2} = 3$. Consequently $u-2=1$, which means $u=3$. In other words, $k=8$ is strong. Conclusion: The positive integer $k$ is strong iff $k \in \{2,3,4,8\}$.A positive integer $k$ is called strong if there exists two positive integers $r$ and $s$ satisfying the Diophantine equation $(1) \;\; (k^2 - 6k + 11)^{r-1} = (2k - 7)^s$. If $k=1$, then $6^{r-1} = (-5)^s$ by equation (1), yielding $5 | 6$. Hence $k=1$ is not strong by contradiction. If $k=2$, then $3^{r-1} = (-3)^s$, which is satisfied when $(r,s)=(3,2)$. Therefore $k=2$ is strong. Assume $r=1$. Then $(2k - 7)^s = 1$ by equation (1), yielding $|2k - 7|=1$. Hence $k \in \{3,4\}$ and $s=2$ satisfies equation (1). Thus $k=3$ and $k=4$ are both strong. Next assume $r \geq 2$ and $k \geq 5$. Hence $2k - 7 \geq 2 \cdot 5 - 7 = 10 - 7 = 3$, implying there exists an odd prime divisor $p$ of $2k-7$. Consequently by equation (1) we obtain $p \mid k^2 - 6k + 11$, which means (since $p$ is odd and $2k \equiv 7 \pmod{p}$) $0 \equiv 4(k^2 - 6k + 11) = (2k)^2 - 12 \cdot (2k) + 44 \equiv 7^2 - 12 \cdot 7 + 44 = 49 - 84 + 44 = 9 \pmod{p}$, yielding $p | 3^2$. Therefore $p=3$, which according to equation (1) yields there are two positive integers $u$ and $v$ s.t. $(2) \;\; (k - 3)^2 + 2 = 3^u$, $(3) \;\; 2k - 7 = 3^v$. Hence by equation (2) $(2k)^2 - 12 \cdot (2k) + 44 = 4 \cdot 3^u$, which combined with equation (3) give us $(3^v + 7)^2 - 12(3^v + 7) + 44 = 4 \cdot 3^u$, i.e. $(4) \;\; 3^{2v} + 2 \cdot 3^v + 9 = 4 \cdot 3^u$, or alternatively $(5) \;\; 3^{2v-2} + 2 \cdot 3^{v-2} + 1 = 4 \cdot 3^{u-2}$. Seeing that $k > 4$, we obtain by equation (2) $3^u = (k - 3)^2 + 2 \geq (5 - 3)^2 + 2 = 2^2 + 2 = 4 + 2 = 6 > 3$, which give us $u \geq 2$. Hence, since $v \geq 1$, we have $3^{v-2} \in \mathbb{N}$, i.e. $v \geq 2$. If $u,v \geq 3$, then $3 | 1$ by equation (5), implying $u=2$ or $v=2$. If $u=2$, then according to equation (4) we obtain $(3^v + 1)^2 = 28$, implying $u \neq 2$. Hence $v=2$, yielding $2k-7=3^v=3^2=9$, i.e. $k=8$. Inserting $v=2$ in equation(5), the result is $4 \cdot 3^{u-2} = 12$, i.e. $3^{u-2} = 3$. Consequently $u-2=1$, which means $u=3$. Inserting $k=8$ in equation (1), we obtain $3^{3(r-1)} = 3^{2s}$, an equation which is satisfied if $r=s=3$. This fact means $k=8$ is strong. Conclusion: The positive integer $k$ is strong iff $k \in \{2,3,4,8\}$.

nice sol @above! we started similarly, but my proof for $k\geq 5$ worked differently.

InternetPerson10 wrote: Determine all positive integers $k$ for which there exist positive integers $r$ and $s$ that satisfy the equation $$(k^2-6k+11)^{r-1}=(2k-7)^{s}.$$ Suppose that $k>=9$ then: $k^{r-1}*3^{r-1}<=[k(k-6)]^{r-1}<LHS=RHS<2^s*k^s$ so $s>=r-1$ which means that $2k-7||k^2-6k+11$ $4k^2-24k+44=7^2-12*7+44=9(mod 2k-7)$ So $2k-7|9$ contradiction because $2k-7>=2*9-7=11$. Now for $k=1,2,3...,7,8$ we get only the values $k=2,3,4,8$