How prove $\deg (f^{m} - g^{n})\ge \frac{mn - m - n}{n}\deg f + 1$? 
Let $m,n\in \mathbb{N}$ be such that $m,n\ge 2$. Let $f,g$ be non-constant polynomials (with real coefficients, or more generally - complex) such that $f^{m} - g^{n}\neq 0.$ How prove $$\deg (f^{m} - g^{n})\ge \frac{mn - m - n}{n}\deg f + 1\ ?$$ 


Observe that the claim is trivial unless $m\deg f=n\deg g$, because otherwise $\deg(f^m-g^n)=\max\{m\deg f,n\deg g\}$. Thus we can assume that this is the case. Given this relation we also have
$$
\frac{mn-m-n}n\deg f=\frac{mn-m-n}m\deg g,
$$
so the claimed inequality is symmetric in $f$ and $g$ in spite of appearance to the contrary.
 A: I really like this question and I was quite disappointed to see it closed first.  Thanks Jyrki for editing.  This post is not an answer (yet), just a longer observation.
We have already seen in the comments that $m\deg(f) = n\deg(g)$ is necessary, so assume this.  The condition on the right hand side can be reformulated as $$\deg(f^m - g^n) \ge m\deg(f) - \deg(f) - \deg(g) + 1$$
This made me think of some sort of linear condition since $m\deg(f) + 1$ is the vector space dimension of the space that $f^m-g^n$ lives in and $\deg(f) + \deg(g) + 2$ is the dimension of the space that is parametrized by picking $f$ and $g$.  Now I can't find the linear conditions and also the dimensions don't quite fit.  
Here is an algebraic condition however: There is a polynomial map on coefficients, that is $\mathbb{C}^{\deg(f) + \deg(g) + 2} \to \mathbb{C}^{m\deg(f)}$ given by $(f,g) \mapsto f^m - g^n$.  To give the simplest example, let $f=ax+b$, $g=cx+d$, and $m=n=2$. Then $$f^m-g^n = (a^2-c^2)x^2 + 2(ab-cd)x + b^2-d^2.$$  The algebraic map is $$(a,b,c,d) \mapsto (a^2-c^2, 2(ab-cd), b^2-d^2).$$ The question is now about points in the image and if their coordinates can vanish. For example, asking if, in the example, the result can be a constant is to ask what are the solutions to $a^2-c^2=2(ab-cd)=0$.  Primary decomposition in Macaulay2 quickly yields that happens only if $f=g$ or $f=-g$ or $a=c=0$.
Maybe somebody more fluent in algebraic geometry can recognize the equations that arise in the general case and finish the proof.
A: If $f,g$ are non-constant coprime polynomials one may use Mason-Stothers theorem. 
We suppose that $m\deg f=n\deg g$.
We have $(f^m-g^n)+g^n=f^m$. Let $t$ be the number of (distinct) zeros of the polynomial $(f^m-g^n)g^nf^m$. Then $$m\deg f= \max(m\deg f,n\deg g,\deg(f^m-g^n))\le t-1.$$ But $t\le\deg f+\deg g+\deg(f^m-g^n)$, and thus we get $$m\deg f\le\deg f+\deg g+\deg(f^m-g^n)-1$$ which is equivalent to $$\deg(f^m-g^n)\ge(m-1)\deg f-\deg g+1.$$
Now it is easily seen that $(m-1)\deg f-\deg g+1= \dfrac{mn - m - n}{n}\deg f + 1$, and we are done.
