$(x^n-x^m)a=(ax^m-4)y^2$ in positive integers How do I find all positive integers $(a,x,y,n,m)$  that satisfy $ a(x^{n}-x^{m}) = (ax^{m}-4) y^{2} $  and $ m\equiv n\pmod{2} $, with $ax$ odd?
 A: Some partial results.
Case 1) $n<m$.
Then  $ax^m<4$. 
The solutions of this inequality are:
$$
a=1, m=1, x=1,2,3,\\
          m\geq2, x=1,       
$$
and
$$
a=2,3, x=1, m\geq 1.
$$
From these we obtain two candidates (i) $x=1$ and (ii) $x=3$. 
(i) Substituting into the original equation we get
$$
0=(a-4)y^2.
$$
Since $y>0$ and $ax^m=a<4$ there is no solution.
(ii) Substituting into the original equation we get
$$
3^n-3=(3-4)y^2.
$$
Since $n<m$ and $n\equiv m \pmod{2}$, thus $n=1$ which gives $y=0$ that is no solution.
Case 2) $n=m$.
Substituting into the original equation we have
$$
(ax^m-4)y^2=0.
$$
Since $y>0$ we get 
$$
ax^m=4.
$$
Since $ax$ is odd there is no solution.
Thus we have justified that $n>m$.
So we can write 
$$
x^m(x^{n-m}-1)a=(ax^m-4)y^2.
$$
Since $a$ is odd and $(a,ax^m-4)=1$, therefore $a|y^2$, and similarly $x^m|y^2$.
Furthermore $n=m+2k$ where $k$ is positive integer. Thus we obtain
$$
ax^m(x^{2k}-1)=(ax^m-4)y^2.
$$
Introducing the new variables $u:=ax^m$, $z:=x^k$ we obtain
$$
u(z^2-1)=(u-4)y^2,
$$
where $u,z$ are odd and $z>1$. Obviously $8|z^2-1$, thus $4|y$. Since 
$$
u(z^2-1)=(u-4)y^2<uy^2,
$$
it follows $z<y$.
A: Brute force finds three solutions so far:  
a=3, x=3, y=12, n=5, m=1
a=1, x=3, y=12, n=6, m=2
a=1, x=9, y=12, n=3, m=1
which are just different ways of saying that $729 - 9 = (9-4)\times12^2.$
