Consider the homogeneous polynomial $R(x, y) = P(x, y) - Q(y, x)$; it can be written as
$$R(x, y) = r_0x^n + r_1x^{n-1}y + \dots + r_kx^{n-k}y^k + \dots + r_{n-1}xy^{n-1} + r_ny^n.$$

By assumption we have $R(k, n - k) = 0$ for $k = 0, 1, 2, \dots, n$. For $k = 0$ and $k = n$ we get $0 = R(0, n) = r_n n^n$ and $0 = R(n, 0) = r_0 n^n$. Therefore,

$r_0 = r_n = 0$ and each term of $R(x, y)$ is divisible by $xy$. If $n = 1$, this means that $R = 0$. For $n \ge 2$ there is a homogeneous polynomial $S(x, y)$ of degree $n-2$ such that $xyS(x, y) = R(x, y)$.
Our aim is to show that $R(x, y) = 0$ and so we can assume $n \ge 2$. From the univariate polynomial $\tilde{S}(x) = S(x, 1)$ we can recover $S$, because
$$S(x, y) = y^n \tilde{S}(x/y).$$
This is so because $S$ is homogeneous and $x^{n-k}y^k = y^n(x/y)^k$. To prove that $R = 0$ it is therefore sufficient to show that $\tilde{S} = 0$.
By assumption we have $S(k, n-k) = 0$ for $k = 1, 2, \dots, n-1$. From $S(k, n-k) = (n-k)^n \tilde{S}(k/(n-k))$ we deduce that the $n-1$ distinct numbers $\frac{1}{n-1}, \frac{2}{n-2}, \dots, \frac{n-2}{2}, \frac{n-1}{1}$ are roots of the polynomial $\tilde{S}(x)$. Because the degree of $\tilde{S}(x)$ is at most the degree of $S(x, y)$, which was equal to $n-2$, this implies that $\tilde{S} = 0$ and so $R = 0$ as well.