Prove that the number 14641 is the fourth power of an integer in any base greater than 6?

In base $6$, $14641$ is not a valid representation of a number, because $6$ is not a valid digit in base $6$.

It is true that $1+4\cdot 6 + 6\cdot 6^2+4\cdot 6^3+6^4$ is still equal to $7^4$. But the proper representation for this number in base $6$ is $15041_6$.

Your question generalizes. The expansion of $(b+1)^n$ is clearly a perfect $n$th power for every base $b$ greater than the largest binomial coefficient (the middle one). When $n > 4$ the middle binomial coefficient has more than one (decimal) digit, so you can't write the expansion using the digits $0, \ldots 9$.

Even more generally, the expansion of $(ab+c)^n$ is a perfect $n$th power when the base $b$ is greater than the maximum of the terms ${n \choose i}a^{i}c^{n-i}$. That's the condition that there be no carries when you calculate using the standard (in the U.S.) multiplication algorithm in base $b$.

Unfortunately, you can't get far when you're limited to base 10 digits. The two digit number powers you can compute before you have to start carrying are $11^4 = 14641$, $12^2 = 144$, $13^2 = 169$, $22^2 = 484$ and the reversals $21^2 = 441$ and $31^2 = 961$.

More examples:

$(110 + 1)^3 = 111^3 = 1367631$, $112^2 = 12544$, ...

$1111^2 = 1234321$, ... , $(11\ldots1)^2 = 12345678987654321$

For homework: find them all.

As a matter of personal preference, I like $b$ to represent the base rather than $a$.

So then $(b + 1)^2 = b^2 + 2b + 1$. And $(b^2 + 2b + 1)^2 = b^4 + 4b^3 + 6b^2 + 4b + 1$.

These facts are true whether $b$ is an ordinary positive integer greater than 1 or a more "exotic" number, like $\sqrt{-2}$. But whether $(b + 1)^4$ gets represented as 14641 in base $b$, that's a slightly different story.

Consider for example $b = 2$. Indeed $3^4 = 2^4 + 4 \times 2^3 + 6 \times 2^2 + 4 \times 2 + 1$. But the problem is that here it turns out that $6 > b^2$ and $4 > b$. Then $4b^3$ requires more than four binary digits to represent, $6b^2$ requires more than three bits and $4b$ requires more than two bits. So the binary representation of 81 is 1010001 rather than 14641.

Some of these problems persist through $b = 6$, because although $4b^3 < b^4$, we still have $6b^3 > b^3$. Then 2401 is 15041 in base 6.

The smallest integer $b$ satisfying $4b^3 < b^4$, $6b^2 < b^3$ and $4b < b^2$ is $b = 7$. Indeed 4096 in base 7 is 14641.