A 4, 4, 1 digits form a three-digit number that is a perfect square in every number system whose base is greater than 4. Let's find a base of a number system in which 441 is a perfect cube, and also a perfect fourth power.
a) Let the base of the number system be denoted by $a$. The three-digit number represented by the digits 4, 4, 1 can be written as:

$$
A=441=4 a^{2}+4 a+1=(2 a+1)^{2}
$$

Thus, $A$ is indeed a perfect square. The condition $a>4$ is necessary only to ensure that the digit 4 is included among the digits of the number system.

b) If $A$ is a perfect cube, then from the previous expression it follows that $(2 a+1)$ itself must be a perfect cube, i.e.,

$$
2 a+1=n^{3}
$$

from which

$$
a=\frac{n^{3}-1}{2}
$$

Since $a$ must be an integer, $n=2 k+1$ and because $a>4$, $k=1,2,3, \ldots$. Thus, $A$ is a perfect cube if the base of the number system is $13,62,171, \ldots$

c) If $A$ is a perfect fourth power, then $(2 a+1)$ must be a perfect square, so

$$
2 a+1=n^{2}
$$

from which

$$
a=\frac{n^{2}-1}{2}
$$

Similarly to the previous case, $n$ is also odd, $n=2 k+1$ and because $a>4$, $k=2,3, \ldots$

Thus, $A$ is a perfect fourth power if the base of the number system is $12,24,40, \ldots$. This completes the solution of the problem.

Note. The problem clearly refers to the number 441, so it was not necessary to examine the numbers 414 and 144, where the former does not even satisfy the condition of the problem that it is a perfect square in any number system with base $a>4$ as some solvers did.