suppose $a^n+b^n$ has an odd prime factor $p$ that doesnt divide $a$ neither $b$. then $p^{k(n+1)}||a^n+b^n$ now we must have that $a^{npq}+b^{npq}$ is a perfect $npq+1$ power for $q$ an odd prime different from $p$, the order of $p$ here is $kn+k+1$ so for big enough $q$ we have $npq>kn+k$, and thus the number is not a perfect $npq+1$ power... So let $p_1,p_2...,p_m$ be the common prime factors of $a$ and $b$, we must then have that: $a^n+b^n$ can only have the prime factors $p_1,p_2...,p_m$ and $2$, now if $n$ is odd we get all primes factors of $a+b$ are a subset of those from $a^n+b^n$ from here, let $d=(a.b)$, then $a+b=2^kd$, but then $a^q+b^q$ for q different from $2$ and $p_i$, has a prime factor different from those of $a+b$, and this new prime doesnt divide $a$ neither $b$...so we get the initial contradiction, unless $a=b=1$ but this case doesnt work.... Please, can someone chek it???? thanks!

there are gaps in the above post.... Until now we just have to say, that $a^n+b^n$ has all prime factos divisors of $2ab$....some cases may be dealed here, but i dont have time right now...

If $a=b$ it is easy... Let $ a\geq b+1 $ and $a^n + b^n = (k_n)^{n+1}$ for $ k_n \in N $ Since $a \geq 2$ we get $ a^{n+1}\geq 2a^n \geq (k_n)^{n+1} + 1$ so $a\geq k_n +1$ As a result there is a $k\in N$ such that $a^n + b^n = (k)^{n+1}$ for infinetely many n. But then $ak^n\geq k^{n+1}\geq a^n$ so $ a \geq (\frac{a}{k})^n \rightarrow \infty$ since $a\geq k+1$ which is a contradiction. The rest follows.

Very nice solution Anto!

Anto wrote: As a result there is a $k\in N$ such that $a^n + b^n = (k)^{n+1}$ for infinetely many n. Why?

Because $a$ is fixed and $k_n \leq a -1$ is bounded for all $n$...