How many integers of $ m $ digits are equal to the sum of the $ m $ -th powers of their digits in base $7$ in the interval $[1, ..., 10 ^ 7]$?

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In a previous question I was checking the following number theory excercise:

The number $1634$ has an interesting property. This 4-digit number satisfies that the sum of the fourth powers of its digits gives the same number. That is, $1 ^ 4 + 6 ^ 4 + 3 ^ 4 + 4 ^ 4$ $=$ $1634$. How many integers of $ m $ digits are equal to the sum of the $ m $ -th powers of their digits in the interval $[1, ..., 10 ^ 7]$? What is the largest of those complies with said property in the same interval?

Through many calculations and help of people here We found $15$ numbers with the respective characteristics. Now I have an extension of that problem:

The above problem can also be considered in other bases. For example, in base $5$, the integer $28$ fulfills this condition. In effect, $28$ in base $5$ is $(103) _5$, and the third powers (length of the integer in base $5$) of his figures is $1 ^ 3$ $0 ^ 3$ $3 ^ 3$ = $1+0+27$ = $28$. How many integers with the previous property are in base $7$ contained in the interval [$1, ..., 10 ^ 7$]? Which one is the biggest?

For the first part of the problem to find the $15$ numbers with the first conditions I noted that for any given exponent only need to check up to some number of digits. For exponent $k$ and $m$ digits the greatest the sum can be is $m9^k$, but an $m$ digit number is at least $10^{m-1}$. Roughly when $m \gt k$ I had $10^{m-1} \gt m9^k$ and I was ok with that $k$. But now I'm dealing with the base change, how do I modify my first thoughts about it based on that? Any help will be really appreciated.

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It works exactly the same when you fix the constants. An $m$ digit number in base $b$ is at least $b^{m-1}$. The greatest the sum can be is $m(b-1)^k$. There will be an $m$ where the sum is guaranteed to be less than the number.