(1).Some restrictions apply. Assume $m>1$ and allow the possibility $d=0.$
(2). Part of the axiomatic definition of $\Bbb N$ is the Principle of Induction: Let $P(n)$ mean that $n$ has property $P,$ which can be any property stated in the formal language of arithmetic. If $P(n)\implies P(n+1)$ for all $n\in \Bbb N$, and if $P(1)$, then $P(n)$ is true for all $n\in \Bbb N.$
From this we can prove that if $P(n)$ for some $n\in \Bbb N$ then there is a least $n\in \Bbb N$ for which $P(n)$ is true.
(3).The Principle of Induction implies that if $m,n\in \Bbb N$ and $m>1$ then there exists $e\in \Bbb N$ for which $n< m^e.$
For if we suppose not, we obtain a contradiction as follows: Let $P(n)$ be "$m^e\leq n$ for all $e \in \Bbb N$". Let $n_0$ be the least $n\in \Bbb N$ with property $P.$
Then $n_0-1\in \Bbb N$ because $n_0\geq m^1=m>1.$
For any $e\in \Bbb N$ we have $e'=e+1\in \Bbb N$ so $m^{e'}\leq n_0,$ so $$m^e=m^{e'}/m\leq n_0/m\leq n_0/2\leq n_0-1.$$ Thus $m^e\leq n_0-1$ for all $e\in \Bbb N, $ and since $n_0-1\in \Bbb N$ we have $P(n_0-1).$ But this contradicts the requirement that no member of $\Bbb N$ which is less than $n_0$ has property $P.$
(4). So let $e_0$ be the $least $ $ e\in \Bbb N$ such that $n<m^e.$ Let $d=e_0-1.$
If $e_0=1$ then $d=0$ and $1=m^0=m^d=m^0=1\leq n <m^{e_0}=m^{d+1}.$
If $e_0>1$ then $d\in \Bbb N$ so $n\geq m^d$ (otherwise $m^{e_0}$ would not be the least positive power of $m$ that is greater than $n.$) So again we have $m^d\leq n<m^{e_0}=m^{d+1}.$
....................Remark:Arithmetic on $\Bbb N$ can be extended to arihmetic on a larger collection $\Bbb N^*$ which has members that are larger than any member of $\Bbb N,$ but the Principle of Induction does not apply to $\Bbb N^*$. So we cannot avoid this principle and still prove that $d$ exists. (Otherwise we could be talking about $m,n\in \Bbb N^*$ and $d\in \{0\}\cup \Bbb N,$ where in some cases, $d$ will not exist.)