Solution:
$k = 3$, $3^{3} + 1 = 28$, $3^{4} + 1 = 82$.

$k = 10$ does not work because $k^{a} + 1$ is a palindrome. If $k > 10$, then for $a < b$ we have $k^{b} + 1 \geq k^{a + 1} + 1 \geq 11k^{a} + 1 \geq 10(k^{a} + 1) + k^{a} - 9 > 10(k^{a} + 1)$. So $k^{b} + 1$ has more digits than $k^{a} + 1$. So we only need to consider $k = 2, 3, \ldots, 9$.

WLOG $a < b$. Suppose $2a < b$. Then $k^{b} + 1 > k^{a}k^{a + 1} > k^{a}(k^{a} + 1)$, so $k^{a} < 10$. But $k^{a} + 1 \geq 10$ (or reversing its digits would not change it). Hence $k^{a} + 1 = 10$, which obviously does not work. Hence $2a \geq b$. So $a \geq b - a$. Hence $k^{b} + 1 > k^{b} - 1 \geq k^{b} - k^{a} + k^{b - a} - 1 = (k^{a} + 1)(k^{b - a} - 1)$. So $k^{b - a} - 1 < 10$. If $k^{b - a} - 1 = 9$, then $k^{b - a} = 10$, so $k = 10$, which we already know does not work. Hence $k^{b - a} - 1 < 9$.

But $(k^{b} + 1) - (k^{a} + 1) = (k^{b - a} - 1)k^{a}$. This must be divisible by 9, because $k^{b} + 1$ and $k^{a} + 1$ have the same digit sum and hence are the same mod 9. Hence $k^{a}$ must be divisible by 9, so $k$ must be 3, 6 or 9.

If $k = 6$ or 9, then $k^{b - a - 1} \geq 5$, so the first digit of $k^{a} + 1$ must be 1 or $(k^{b - a} - 1)(k^{a} + 1)$ would have more digits than $k^{a} + 1$. But that means the last digit of $k^{b} + 1$ is 1 and hence the last digit of $k^{b}$ is 0, which is impossible. So $k = 3$. It is easy to check that there is a solution for $k = 3$.