Let us represent a number $x$ as a string of "digits" where each $n^{th}$ digit represents the power that you raise the $n^{th}$ prime to in its prime factorization
For example, $15=5^13^12^0$ so we would write $15=110$. In the event that a prime power is larger than $9$, we would just specify that the specific digit is larger. For example
$$ 885735=5^13^{11}2^0\rightarrow 1(11)0 $$
Now we can see that $12=3^12^2$ so we would write $12=12$.
Is it the case that $12$ is the only number that is the same in both of these representations? I wrote some code to check the first $10^6$ cases and found that we only have $12$.
Clearly we cannot have $x$ be an odd number if we want the representations to be the same, but outside of simple cases like this I cannot think of a way to trim the cases
This is not a complete answer, but too long to fit in a comment. This shows that if $k\geq100$, then $k$ must be divisible by the cube of a prime. There is plenty of room for improvement of the argument, to restrict the factorization of $k$ further.
Let $p_n$ denote the $n$-th prime number, with $p_1=2$. For any positive integer $k$ with prime factorization $k=\prod p_n^{e_n}$, define $$f(k):=\sum e_n10^n.$$ Let $k$ be a positive integer such that $e_n\leq9$ for all $n$, and $f(k)=k$. If $k$ is odd then $e_1=0$ and so $f(k)$ is a multiple of $10$, contradicting the fact that $f(k)=k$ is odd. Then $k$ is even and so $e_1>0$. Then $e_1\in\{2,4,6,8\}$ and the final digit of $k$ is not $0$, so $k$ is not divisible by $5$ and so $e_3=0$. Moreover, if $e_1\in\{2,6\}$ then $e_2$ is odd, otherwise $e_2$ is even.
The largest prime divisor of $k$ is $p_m$ where $m=\lceil\log_{10}k\rceil$, and so $$k=\prod_{n=1}^mp_n^{e_n}\leq\prod_{n=1}^mp_n^9=(p_m\#)^9,$$ where $p_m\#$ denotes the primorial function, which is by definition the product of the first $m$ prime numbers. If $m\geq3$ we can improve this slightly by using the fact that $k$ is not divisible by $5$, to find $$k\leq\frac{1}{5^9}(p_m\#)^9.$$
In fact if $e$ is the largest exponent of a prime power in the factorization of $k$, we find in the same way that $$k\leq\frac{1}{5^e}(p_m\#)^e,$$ where still $m=\lceil\log_{10}k\rceil$. Then $m<1+\log_{10}k$, and from the upper bound $p_m\#\leq2.763^m$ (see the wikipedia page) it follows that $$k\leq\frac{1}{5^e}(p_m\#)^e<2.763^{e(1+\log_{10}k)}=\left(\frac{2.763}{5}\right)^e\cdot k^{e\log_{10}2.763}.$$ This shows that $e\log_{10}2.763>1$ and hence that $e\geq3$. That is to say $k$ is divisibile by the cube of a prime.