Are there nontrivial rational solutions of $x^{n-m}=(1+t^m)/(1+t^n)$? Let $n-m \ge 2$ be two fixed natural numbers. Are there any nontrivial rational solutions of the equation $$x^{n-m}=(1+t^m)/(1+t^n)$$ for $x$ and $t$? As particular cases the 
 rational solutions of the equations $x^2=(1+t^2)/(1+t^4)$ and $x^2=(1+t^3)/(1+t^5)$ will be interesting.
 A: Of course, t=1 is a solution, but we count that as trivial and move on.
As polynomials in t, we have a ratio of (some products of) cyclotomic polynomials, and the gcd for most choices will be $1+ t^d$ when $d$ is  gcd of m and n and both m/d and n/d are odd. Removing that as a common factor, it is unlikely that what remains has any chance of yielding a comparable power of a rational number.
For m=2 and n=4, we consider t in reduced form as r/s, and look for coprime integers r and s such that the ratio of $(r^2 + s^2)s^2$ to $r^4 + s^4$ is the ratio of two integer squares. Since r and s are coprime, the only common divisor of the two terms can be 1 or 2. Fermat says no for 1, and Max says no for 2.
For m=3 and n=5, we can divide out by (1+t) and note that the two polynomials are coprime. Again picking r and s coprime, we need $u=r^2 -rs + s^2$ to be a square, and the denominator $D$ to be a square.  We then find $D-u^2$ is $rs(u-rs)$ with $u-rs$ also being a square. We also have $D-r^2s^2$ is $r ^2 + s^2$ times the square $u$.  While these conditions do not lead to a contradiction, I strongly suspect a similar manipulation will.
Granville recently proved a result regarding certain sequences having primitive prime factors to an odd power.  I suspect that can be used to tackle the case n-m is even. I will include the reference later in a comment.
Gerhard "Sorry, The Contradiction Went Elsewhere" Paseman, 2018.02.01.
