Cute! Let's assume $p\nmid x, p\nmid y$ as if it did, we can take the common $p$ multiple. Since $x^n+y^n=p^k\implies p|x+y.$ Infact we get $x+y=p^l$ for some positive $l.$ So using LTE, we get $v_p(x^n+y^n)=v_p(x+y)+v_p(n)=l+v_p(n)=k\implies v_p(n)=k-l.$ Now let $n=p^{k-l}\cdot \alpha,~~\gcd(\alpha,p)=1.$ Then we know that $x^{\alpha}+y^{\alpha}|x^n+y^n\implies x^{\alpha}+y^{\alpha}$ a perfect power of $p.$ Now again using LTE, we get $v_p(x^{\alpha}+y^{\alpha})=v_p(x+y)+v_p(\alpha)=l.$ But we have $x+y=p^l\implies \alpha=1.$ So $n$ is a power of $p.$

jelena_ivanchic wrote: Cute! Let's assume $p\nmid x, p\nmid y$ as if it did, we can take the common $p$ multiple. Since $x^n+y^n=p^k\implies p|x+y.$ Infact we get $x+y=p^l$ for some positive $l.$ So using LTE, we get $v_p(x^n+y^n)=v_p(x+y)+v_p(n)=l+v_p(n)=k\implies v_p(n)=k-l.$ Now let $n=p^{k-l}\cdot \alpha,~~\gcd(\alpha,p)=1.$ Then we know that $x^{\alpha}+y^{\alpha}|x^n+y^n\implies x^{\alpha}+y^{\alpha}$ a perfect power of $p.$ Now again using LTE, we get $v_p(x^{\alpha}+y^{\alpha})=v_p(x+y)+v_p(\alpha)=l.$ But we have $x+y=p^l\implies \alpha=1.$ So $n$ is a power of $p.$ I might be overlooking something but how does $v_p(\alpha) = 0$ imply $\alpha = 1$? Also can you explain $v_p(x^{\alpha}+y^{\alpha})=v_p(x+y)+v_p(\alpha)=l$ further?