If $x = y$, then $n = 1$ works, since $x^2 + y^2$ is an even number greater than $2$.

Now assume $x \neq y$. In this case, we will show that for some $n$, the number $x^{2^n} + y^{2^n}$ is divisible by $257$ and not equal to $257$. Then it will be composite, as required.

Suppose $x^{2^n} + y^{2^n} = 257$. Let $a = x^{2^{n-1}}$, $b = y^{2^{n-1}}$. Then $a^2 + b^2 = 257$, and if $a \ge b$, then $a = 16$, $b = 1$ (the cases $a = 12, 13, 14, 15$ are easily checked; if $a \le 11$, then $b^2 \le a^2 < 257/2$, which is impossible). But this contradicts the condition $x, y > 1$. Thus, $x^{2^n} + y^{2^n} \neq 257$.

It remains to check only that for some $n$, the number $x^{2^n} + y^{2^n}$ is divisible by $257$.

Since $y$ is not divisible by the prime number $257$, there exists a natural $q$ such that $x \equiv qy \pmod{257}$. Since $x \neq y$ and $0 < x + y < 257$, we get $q \neq \pm 1 \pmod{257}$. Also, $q \neq 0 \pmod{257}$, since $x$ is not divisible by $257$.

Since $257$ is prime, by Fermat's little theorem, the number $q^{256} - 1 = (q-1)(q+1)(q^2+1)(q^{2^2}+1) \dots (q^{2^7} + 1)$ is divisible by $257$. The first two factors are not divisible by $257$, so for some $n \in \{1, 2, \dots, 7\}$, the number $q^{2^n} + 1$ is divisible by $257$. But then the number $x^{2^n} + y^{2^n} \equiv y^{2^n} (q^{2^n} + 1) \pmod{257}$ is also divisible by $257$.