Isn't there any divisor $k$ of $n^4$ such that $n^2-nI did some experiment with my Python script to find a number which could divide $n^4$ in this interval ($n^2-n$,  $n^2$).
I watched the form of prim factors of the numbers in this ($n^2-n$,  $n^2$) interval, but I hadn't was any idea for proofing it.
Can anyone to help with some start idea to proof that if $n^2-n < k < n^2$ then $k$ can't divide $n^4$.

Answer to the Question
Let $k=n^2-a$. Since $n^2-a\mid n^4$ and
$$
\frac{n^4}{n^2-a}=n^2+a+\frac{a^2}{n^2-a}
$$
we must have $d=\frac{a^2}{n^2-a}\in\mathbb{Z}$. However, since $\color{#C00}{1\le a\le n-1}$ and $\frac{a^2}{n^2-a}$ is increasing in $a$,
$$
0\lt\overbrace{\ \frac1{n^2-1}\ }^{\large\color{#C00}{a=1}}\le d\le\overbrace{\frac{n^2-2n+1}{n^2-n+1}}^{\large\color{#C00}{a=n-1}}\lt1
$$
which is impossible because there is no integer between $0$ and $1$.

Bounding the Greatest Factor of $\boldsymbol{n^4}$ less than $\boldsymbol{n^2}$
Case: $\boldsymbol{d=1}$
Since $n^2=a^2+a$, we have $n-1\lt a\lt n$, which is impossible.
Case: $\boldsymbol{d=2}$
Since $n^2=\frac{a(a+2)}2$, $\sqrt2\,n-1\lt a\lt\sqrt2\,n$. In fact, $2n^2+1=(a+1)^2$. That means that
$$
\left(\frac{a+1}n\right)^2=2+\frac1{n^2}\lt\left(\sqrt2+\frac1{2n^2}\right)^2
$$
Which forces $\frac{a+1}n$ to be a continued fraction overestimate of $\sqrt2$ . This gives the first pairs $(n,a)$ to be
$$
\{(0,0),(2,2),(12,16),(70,98),(408,576),(2378,3362),(13860,19600)\}
$$
The recursion for $n_k$ is $n_k=6n_{k-1}-n_{k-2}$ and $a_k=\left\lfloor\sqrt2\,n\right\rfloor$.
In any case, we have

The largest factor of $n^4$ less than $n^2$ must be less than $n^2-\sqrt2\,n+1$ and there are an infinite number of $n$ so that the largest factor of $n^4$ less than $n^2$ is greater than $n^2-\sqrt2\,n$.

