TITLE: A diophantine equation with unknown exponents: $ 4^n + 5^n = 7^m + 2^m $
QUESTION [10 upvotes]: A friend proposed to me the following problem

Find all naturals $n$ and $m$ such that $ 4^n + 5^n = 7^m + 2^m $

I noticed that no solutions exist such that $m > n$ since then LHS < RHS.
When $m=n$, it seems that the only solutions are $(0,0)$ and $(1,1)$ 
The remaining case seems more challenging $ m < n$. 
I haven't achieved any significant progress yet. 
Would you please share any advice, suggestion, thanks in advance.

REPLY [8 votes]: Claim: all solutions satisfy $m \le 1$.
Assume that there is a solution with $m \ge 2$. Looking at the equation modulo $4$ we get $$1 \equiv 7^m$$
thus, $m$ is even.
Now, looking at the equation modulo $5$, we get
$$4^n \equiv 2\cdot 2^m$$
thus $2^{m+1} \equiv 4$ or $2^{m+1} \equiv 1$ mod $5$. but this is not possible for even $m$, since $2$ raised to an odd power can only be congruent to $2$ or $3$ mod $5$.  
So the only solutions are $(0,0)$ and $(1,1)$.

REPLY [3 votes]: Case $m=n$. By the Mean Value Theorem, there is $s\in (2,4)$ and there is $t\in (5,7)$ such that
$$2ns^{n-1}=ns^{n-1}(4-2)=4^n-2^n=7^n-5^n=f'(t)=nt^{n-1}(7-5)=2nt^{n-1}$$
where $f(x)=x^n$. Now if $n>1$ then $2nt^{n-1}>2ns^{n-1}$ and it follows that there are no solutions for $n>1$.