Did Russians uplift indexes?

Note that if $k$ is even then we have $3^n$ as sum of squares. Bashing through quadratic residues modulo $3$ then tells us that both $x$ and $y$ must be multiples of $3$ , contradicting $\gcd(x,y)=1$ .

Now suppose $k$ is odd. Then $x+y\mid x^k+y^k$ . By Zsigmondy's theorem $x^k+y^k$ contains at least one prime factor that doesn't divide $x+y$ . That means $x^k+y^k$ has at least $2$ prime factors, contradicting the fact that it's a prime power.

However, note that Zsigmondy's theorem has an exception $2^3+1^3$ . We substitute $x=2, y=1,k=3$ then solving gives $n=2$ , a positive integer. Hence $n=\fbox{2}$ is the only solution.

Simple if you know the following facts.

Fact $1$ ) Zsigmondy's theorem

Fact $2$ ) A positive integral power of $3$ can not be the sum of squares of two positive integers.

Because of fact $2$ , we can get rid of all even $k$ 's.

Because of fact $1$ , we cam get rid of all odd $k$ 's with the exception of $2^3+1^3$ which turns out to be a power of $3$ .

So, $3^2=2^3+1^3$ is the only possible answer and the sum of all $n$ is just $\boxed{2}$ .