Suppose $n$ is odd and for the sake of contradiction assume exists such $x,y $. Let $p $ be an odd prime divisor of $x+y $ so by $LTE $ we have $$\nu_p (x^n+y^n)=\nu_p (x+y)+\nu_p (n)$$ But since $n $ is square-free we know $\nu_p (n)\le 1$. Let $\nu_p (x+y)=k $ so it follows $$3k\le k+\nu_p (n)<k+2$$So $2k <2$ contradiction. This implies $x+y=2^l $ for some integer $l$ and since $\gcd (x,y)=1$ it follows $x,y $ are both odd. So $$\nu_2 (x^n+y^n)=l+\nu_2 (x-y)-1\ge 3l$$Implies $\nu_2 (x-y)=\nu_2 (2 (2^{l-1}-y))=1\ge 2l+1$ contradiction since $l\ge 1$. If $n $ is even let $n=2k $ where $k $ is odd. Let $p $ be an odd prime divisor of $x+y $. So $$\nu_p ((x^2)^k+(y^2)^k)=\nu_p (x^2+y^2)+\nu_p (k) $$It follows $2+\nu_p (x^2+y^2)=2+\nu_p ((x+y)^2-2xy)>3\nu_p (x+y) $ but this can't be true. So $x+y=2^l $ for some integer $l$ so $x,y $ are both odd. $$\nu_2 (x^n+y^n)=\nu_2 (x^2+y^2)+\nu_2 (x^2-y^2)-1$$ So it follows $$\nu_2 (x^2+y^2)+\nu_2 (x-y)>2l $$ But $x^2+y^2=2^{2l}-2xy=2 (2^{2l-1}-xy) $ and $x-y=2 (2^{l-1}-y) $ so $$2=\nu_2 (x^2+y^2)+\nu_2 (x-y)>2l $$Contradiction since $l\ge 1$. Note: In case $n$ even when I said "This can't be true" follows $$\nu_p ((x+y)^2-2xy)=0$$ Because $p\mid x+y $ but $p\nmid x,y $.

Romania TST 1994

$n$ being odd can be easily dealt with LTE with two cases. Now,if $n$ is even,then consider an odd prime factor $p$ of $x+y$.It can't divide $y$,so we get a contradiction. But if $x+y$ is a power of $2$ ,then mod $4$ gives a contradiction as $x,y$ is odd.

This fails when $n$ is even since we must have $x+y \mid x^n+y^n$ but $x^n+y^n \equiv{2x^n} \pmod{x+y}$ in that case. Therefore $n$ must be odd. Let $p$ be an odd prime. If $\exists \hspace{.01 cm} p \mid a+b$, we might have $p \mid a,b$ which is not possible, so $p \nmid a,b$. Therefore, LTE guarantees that, $$3 \cdot \nu_p(x+y) \leq \nu_p(x^n+y^n)=\nu_p(x+y)+\nu_p(n) \leq \nu_p(x+y) +1$$since $n$ is square free, contradiction. So it follows that $x+y$ must be a power of $2$. However, note that $$x^n+y^n=(x+y)(x^{n-1}- \dots+y^{n-1})$$The number of terms in the second bracket is $n$ which is odd, so $2 \nmid \frac{x^n+y^n}{x+y}$. It implies that $$\nu_2(x^n+y^n)=\nu_2(x+y) \geq 3\nu_2(x+y)$$, a clear contradiction $\blacksquare$

handwritten solution image attached

Note that $x \equiv - y \mod{x + y}$, so if $n$ even $x^n \equiv y^n \mod{x + y} \implies x^n + y^n \equiv 2y^n \mod{x + y}$, and since $(x, y) = 1$, we have $x + y \nmid 2y^n$, a contradiction. So, assume that $n$ is odd. Consider a prime $q$ such that $q \mid x + y$, and $q$ is odd. We have $v_q(x^n + y^n) = v_q(x + y) + v_q(n) \leq v_q(x + y) + 1$, since $n$ is squarefree. However, $v_q((x + y)^3) = 3v_q(x + y) > v_q(x + y) + 1$, a contradiction. If $q = 2$, we have $v_2(x^n + y^n) = v_2(x + y) < 3v_2(x + y)$, a contradiction yet again. Hence, we are done.

Assume that $(x+y)^3 \mid x^n+y^n$ for some $x,y \in \mathbb{Z}^+$, $\text{gcd}(x,y)=1$. Note that $x+y=2 \implies x=y=1 \implies 8 \mid 2$ which is absurd. If $n$ is even, then $x+y \mid x^n-y^n \implies x+y \mid 2y^n$. If $x+y=2^k$ with $k \geq 2$, then $2 \mid y^n \implies 2 \mid y \implies 2 \mid x$, a contradiction. Otherwise let $p$ be an odd prime dividing $x+y$. Then $p \mid 2y^n \implies p \mid y \implies p \mid x$, which is also a contradiction. So $n$ is odd. Again split into two cases. If $x+y=2^k$ with $k \geq 2$, then since $2 \nmid x,y$, by LTE we have $3k \leq \nu_2(x^n+y^n)=\nu_2(x+y)=k$, a contradiction. Otherwise let $p$ be an odd prime dividing $x+y$, then $3\nu_p(x+y) \leq \nu_p(x^n+y^n)=\nu_p(x+y)+\nu_p(n) \leq \nu_p(x+y)+1 \implies \nu_p(x+y)=0$ since $n$ is squarefree. This is also a contradiction. Hence proved.