Determine all $P(x) \in \mathbb R[x]$ such that $P(x^2) + x\big(mP(x) + nP(-x)\big) = \big(P(x)\big)^2 + (m - n)x^2$, $\forall x \in \mathbb R$.

I think you got wrong in the first step.Replacing $x$ by $−x$, we have indeed $P(x^2) - x[mP(-x) + nP(x)] = [P(-x)]^2 + (m - n)x^2, \forall x \in \mathbb R$ but not $P(x^2) + x[mP(-x) + nP(x)] = [P(-x)]^2 + (m - n)x^2, \forall x \in \mathbb R$.

So after substracting we should get $x(m + n)[P(x) + P(-x)] = P(x)^2 - P(-x)^2$.

That's $P(x) + P(-x) = 0$ or $P(x) - P(-x) = x(m + n)$.

In the former case try writing $P(x) = xF(x^2)$, and the latter $P(x) = F(x^2) + \frac{m+n}{2}x$, where $F[x]\in \mathbb R[x]$