= Solution
Put $u=x-y$ and $v=x+y$. The square becomes $(-1,1)^2$, with $x=(u+v)/2$, $y=(v-u)/2$ and absolute <Jacobian determinant> $1/2$. On its support the denominator $v+2$ exceeds one and $|u|<1$, so the nonnegative integrand is bounded by one on a set of finite area; it is therefore <Lebesgue integrable>. The <change of variables formula> and <Fubini's theorem> give
$$
\int_{\mathbb R^2}f^{2n}
=\frac12\int_{-1}^1u^{2n}\,du\int_{-1}^1(v+2)^{-2n}\,dv
=\boxed{\frac{1-3^{1-2n}}{(2n-1)(2n+1)}}.
$$
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