<s>[INST] Prove that $\frac{a^n-1}{b^n-1}$ and $\frac{a^{n+1}-1}{b^{n+1}-1}$ can't both be prime. 
Prove that $$\frac{a^n-1}{b^n-1} \ \text{and} \ \frac{a^{n+1}-1}{b^{n+1}-1}$$ cannot both be prime ($a>b>1,n\ge 2$).  

Clearly $(a^n-1,a^{n+1}-1)=a-1$ and $(b^n-1,b^{n+1}-1)=b-1$.  
Clearly $b-1\mid a-1$ (since $b-1\mid a^n-1,a^{n+1}-1$), so $a=bk-k+1$ for some $k\in\mathbb Z_{\ge 2}$.
A hint would probably be enough. I haven't found an approach I could use.
 [/INST] Let $\let\v\nu\let\geq\geqslant\v_p(n)$ denote the exponent of the prime $p$ in the factorisation of $n\in\mathbb N$.

Lemma (Lifting The Exponent)  If $p$ is odd, $p\nmid a$ and $p\mid a-b$, then $$\v_p(a^n-b^n)=\v_p(a-b)+\v_p(n).$$
  Proof. See ProofWiki.

We already have $b-1\mid a-1$. Let $r$ be any prime divisor of $b-1$ and thus of $a-1$. If $r$ is odd, then (by the lemma)
$$\v_r\left(\frac{a^n-1}{b^n-1}\right)=\v_r(a-1)-\v_r(b-1).$$
Because $\frac{a^n-1}{b^n-1}=p$ is prime, this shows that $\v_r(a-1)=\v_r(b-1)$ for all primes $r\mid b-1$ with $r\neq2$ and $r\neq p$. Hence $a-1=2^kp^xg(b-1)$ for some $k,x$ and $g$, $\gcd(g,b-1)=1$.
We can do the same for $\frac{a^{n+1}-1}{b^{n+1}-1}=q$, giving $a-1=2^lq^yh(b-1)$ for some $l,y$ and $h$, $\gcd(h,b-1)=1$.
Note that $a-1=2^lq^yh(b-1)$ uniquely determines $q$ if $q\mid b-1$ and $\gcd(n,b-1)=1$:


*

*If $q$ is odd, it is the unique odd prime $r\mid b-1$ with $\v_r(a-1)>\v_r(b-1)$

*$q=2$ iff there is no odd prime $r\mid b-1$ with $\v_r(a-1)>\v_r(b-1)$


So we actually have that $p=q$:
$$\frac{a^{n+1}-1}{b^{n+1}-1}=\frac{a^n-1}{b^n-1}$$
hence
$$a^{n+1}b^n-a^{n+1}-b^n=b^{n+1}a^n-b^{n+1}-a^n$$
$$\color{darkgreen}{(a-b)}\color{blue}{(a^nb^n+\frac{a^n-b^n}{a-b}-\frac{a^{n+1}-b^{n+1}}{a-b})}=0$$
but, because $b^n\geq2^n\geq n+1$,
$$\color{blue}{a^nb^n+\frac{a^n-b^n}{a-b}}\geq (n+1)a^n\color{blue}>a^n+a^{n-1}b+\cdots+ab^{n-1}+b^n=\color{blue}{\frac{a^{n+1}-b^{n+1}}{a-b}}$$
contradicting $\color{darkgreen}{a\neq b}$.
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