Suppose that $P_n \mid Q_n$, so that $Q_n(x, y, z) = R_n(x, y, z)P_n(x, y, z)$. Define $p_n(x) = P_n(x, 0, -1)$ and $q_n(x) = Q_n(x, 0, -1)$. Then, we have $q_n = R_n(x, 0, -1)p_n$, so that $p_n \mid q_n$ as real polynomials in $x$. Furthermore, $p_n$ and $q_n$ both have integer coefficients, and $p_n$ is monic. Hence, $R_n(x, 0, -1)$ has integer coefficients as a polynomial in $x$, and in particular, it evaluates to an integer whenever $x$ is an integer. Thus, for any integer $a$, we have $p_n(a) \mid q_n(a)$ as integers.

Plugging in $a = 1$, we find that $2^{2n+1} + 1 \mid (2^{2n} + 2)^{2n}$. Now, suppose $p$ is a prime such that $p \mid 2^{2n+1} + 1$. Then $p \mid 2^{2n} + 2$, and so
$$0 \equiv 2^{2n+1} + 1 - 2(2^{2n} + 2) \equiv -3 \pmod{p}.$$
Therefore, $p = 3$, which means that $2^{2n+1} + 1 = 3^k$ for some $k$.

Note that the left hand side of this equation is always 1 modulo 4, so $k$ must be even. Letting $k = 2m$, we have $2^{2n+1} = (3^m + 1)(3^m - 1)$. It follows that $3^m + 1$ and $3^m - 1$ are both powers of 2, whence $m = 1$ and $n = 1$. Thus, $n = 1$ is the only solution.

We give an alternate way to show that $P_n$ does not divide $Q_n$ for $n \ge 2$. Notice that for any complex $r \ne -1$, we have
$$P_n \left(1, 0, \frac{1}{r+1}\right) = \frac{1}{(r+1)^{2n}} + \frac{r^{2n}}{(r+1)^{4n}} + \frac{r^{2n}}{(r+1)^{2n}} = \frac{(r^{2n}+1)(r+1)^{2n} + r^{2n}}{(r+1)^{4n}}$$
and
$$Q_n \left(1, 0, \frac{1}{r+1}\right) = \left[1 + \frac{1}{(r+1)^{2n}} + \frac{r^{2n}}{(r+1)^{2n}}\right]^{2n} = \left[\frac{(r+1)^{2n} + (r^{2n}+1)}{(r+1)^{2n}}\right]^{2n}.$$
It suffices for us to find an $r$ such that $P_n(1, 0, \frac{1}{r+1}) = 0$, but $Q_n(1, 0, \frac{1}{r+1}) \ne 0$. Set $F_n(x) = (x^{2n} + 1)(x+1)^{2n} + x^{2n}$, so that $P_n(1, 0, \frac{1}{r+1}) = 0$ if and only if $F_n(r) = 0$. In these terms, it suffices to find an $r$ so that $F_n(r) = 0$ and
$$(r + r^2)^{2n} - 1 = F_n(r) + ((r + r^2)^{2n} - 1) = (r + 1)^{2n} + (r^{2n} + 1) \neq 0.$$
Suppose for the sake of contradiction that no such $r$ existed. Then, for any root $r$ of $F_n$, we have that $r^{2n}(r+1)^{2n} = 1$, implying that
$$r^{2n}F_n(r) = r^{2n}[(r+1)^{2n} + r^{2n} + 1] = r^{4n} + r^{2n} + 1 = 0.$$
Thus, $r^{2n}$ is a primitive third root of unity, so $1+r^{2n}$ is the negation of a primitive third root of unity. In particular, both $r^{2n}$ and $1+r^{2n}$ have unit norm. Now, because $F_n(r) = 0$, we have
$$(1+r)^{2n} = \frac{r^{2n}}{1+r^{2n}},$$
which implies that both $1+r$ and $r$ have unit norm and hence that $r$ is itself a primitive third root of unity. Therefore, any root of $F_n(x)$ is a primitive root third root of unity, so we must have $F_n(x) = c(x^2+x+1)^{2n}$ for some non-zero constant $c$. Multiplying both sides by $(x-1)^{2n}$, we obtain
$$c(x^3 - 1)^{2n} = (x-1)^{2n} F_n(x) = (x^2 - 1)^{2n}(x^{2n} + 1) + x^{2n}(x-1)^{2n}.$$
The $x^2$ coefficient of the polynomial on the left is 0, while the $x^2$ coefficient of the polynomial on the right is $-2n$, a contradiction. Therefore, we may find an $r$ with the desired property, as needed.