Let $$P(x)=(x-n^2)Q(x)+2022=(x-n^2)(a_0+a_1x+\ldots+a_dx^d)+2022$$for integers $a_0, a_1, \ldots, a_d$ and let $r=\frac{p} {q}$ for $(p, q)=1$. The second equation $P(r^2)=2024$ rewrites as $$(p^2-q^2n^2)(a_0q^{2d}+a_1p^2q^{2d-2}+\ldots+a_dp^{2d})=2q^{2d+2}.$$Since the first expression is coprime with $q$, this implies that $p^2-q^2n^2 \mid 2$, which is impossible since there aren't two positive perfect squares with difference less than $3$.