Equal length powers

Answer: 3 (1, 2 and 4)

Listing up from 1

$n$	$n^2$	$n^3$	Same number of digits?
1	1	1	yes
2	4	8	yes
3	9	27	no
4	16	64	yes
5	25	125	no
...	2-digit	3-digit	no
10	100	1000	no
...	3-digit	4-digit	no

When $n\gt$10, multiplying by $n$ will increase the number of digits, so $n^3$ (which is $n^2\times n$) will have more digits than $n^2.$

Counting down from 10

When $n\ge$10, multiplying by $n$ will increase the number of digits, so $n^3$ (which is $n^2\times n$) will have more digits than $n^2.$

Numbers less than 10 have squares less than 100 (1 or 2 digits).

Cubes less than 100: 4$^3$ = 64, 5$^3$ = 125

1, 2, 3 and 4 have cubes less than 100

1, 2, 3 have 1-digit squares but 4$2$ = 16 also has 2 digits.

Cubes less than 10: 2$^3$ = 8, 3$^3$ = 27

1 and 2 have cubes and squares less than 10 (1 digit)

So 1, 2 and 4 are the only possibilites.