I'm looking for a set of conditions and maybe a proof of said conditions for the thought proposed in the title. It seems to me that what was stated in the title always is true when N is not a perfect square and N < m, but I can find instances that work for perfect squares N, e.g. m = 12, N = 4.
To be clear, I'm asking how to generalize $\,3\mid m^2\Rightarrow 3\mid m\,$ asked in this question.
It is $(3)$ in the chracterizations of a squarefree integer $q\,$ below.
Theorem $\ $ Let $\rm\ 0 \ne q\in \mathbb Z\:.\ \ $ The following are equivalent.
$(1)\rm\quad\ \ \ \, n^2\,|\ q\ \ \Rightarrow\ \ n\ |\ 1\qquad\ $ for all $\rm\:\ n\in \mathbb Z $
$(2)\rm\quad\ \ \ \, n^2\, |\, qm^2 \!\Rightarrow n\ |\ m\qquad\! $ for all $\rm\: \ n,m\in \mathbb Z$
$(3)\rm\qquad\ q\ |\ n^2\ \Rightarrow\ q\ |\ n\qquad\ $ for all $\rm\:\ n\in \mathbb Z $
$(4)\rm\qquad\ q\ |\ n^k\ \Rightarrow\ q\ |\ n\qquad\ $ for all $\rm\:\ n\in \mathbb Z,\ k\in \mathbb N $
$(5)\rm\quad\:\ \: q^q\ |\ n^n\ \Rightarrow\ q\ |\ n\qquad\ $ for all $\rm\:\ n\in \mathbb N,\ $ for $\rm\ q > 0 $
Proof $\ \: (1\Rightarrow 2)\rm\:\ \ $ Canceling $\rm\:(n,m)^2\:$ from LHS of $(2)\:$ we may assume w.l.o.g. that $\rm\:(n,m)\:=\:1.\ $ By $ $ Euclid's Lemma $\rm\: n^2\, |\, qm^2\: \Rightarrow\ n^2\: |\: q\ \Rightarrow\ n\:|\:1\ \Rightarrow\ n\:|\:m$
$(2\Rightarrow 3)\rm\quad q\ |\ n^2\ \Rightarrow\ q^2\ |\ qn^2\ \Rightarrow\ q\ |\ n $
$(3\Rightarrow 4)\rm\quad k \ge 2\ \Rightarrow\ k \le 2\:(k-1)\ $ so $\rm\:\ q\ |\ n^k\ |\ (n^{k-1})^2\ \Rightarrow\ q\ |\ n^{k-1}\:\ldots\:\Rightarrow\ q\ |\ n$
$(4\Rightarrow 5)\rm\quad q\ |\ q^q\ |\ n^n\ \Rightarrow\ q\ |\ n $
$(5\Rightarrow 1)\:$ via $\:\lnot\: 1\Rightarrow\lnot\: 5.\ $ By $\rm\:\lnot 1,\,\ q\: =\: ab^2,\:\ b\nmid 1.\:$ Put $\rm\ n = abc\:$ for $\rm\:c\:$ as below.
$\rm\qquad\ \ \ q\ |\ (ab)^2\ \Rightarrow\ q^{\:q}\ |\ (ab)^{2\:q}\ |\ (abc)^{abc}\! = n^n\quad\ \ for\ all \:\ c\:\ with\ \ abc > 2\:q $
Since $\rm\ b\nmid 1\ \Rightarrow\ q\nmid ab,\:$ we may choose $\rm\:c\:$ so that also $\rm\ q\nmid abc,\ $ e.g. $\rm\:\ c\equiv 1\,\ (mod\ q)$
(I assume $m$ integer and $N$ non-zero integer.)
The proposed statement is true for all $m\in\mathbb{Z}$ if and only if $N$ is square-free (this is obvious). If $N=k^2l$ with $l$ square-free and $k\neq\pm1$ then $m=kl$ is a counterexample.
More precisely: We have that $N\mid m^2\implies N\mid m$ for all $m\in\mathbb{Z}$ if and only if $N$ is a square-free integer.
Example for $N$ not a perfect square: $N=12$, $m=6$.
If you wanted $m>N$: $N=12$, $m=18$.
A general statement is not really interesting: We have that $N\mid m^2\implies N\mid m$ if and only if $N\mid m$ or $N\nmid m^2$. This is not quite useful, is it?
