By the Binomial theorem, one has that $$(a+b)^n = a^n + \binom{n}{1} \ a^{n-1}b + \binom{n}{2}\ a^{n-2}b^2 + \dots + b^n$$
Let us suppose that $d\mid (a+b)$, where $\mid$ means "divides", and such that $d>1$, $n>1$ and $\gcd(a,b)=1$
Could it be proved that $d^n\nmid a^n+b^n$?
$$2 \mid 3 + 5; \quad 2^3 \mid 3^3 + 5^3.$$ Thus, $(a, b, d, n) = (3, 5, 2, 3)$ is a counterexample.