$P(x)$ is a biquadratic polynomial, so it's very easy to find the roots.Let $x^4 - 2(n+p)x^2 + (n-p)^2 = 0$. Then, if $y = x^2$, we have $$y^2 - 2(n+p)y + (n-p)^2 = 0 \Rightarrow y = (n+p) \pm 2\sqrt{np}.$$Hence, if $\alpha = n+p+2\sqrt{np}$ and $\beta = n+p - 2\sqrt{np}$, we have that $\sqrt{\alpha}$, $\sqrt{\beta}$, $-\sqrt{\alpha}$ and $-\sqrt{\beta}$ are all roots for $P(x)$. Now, since $P(x)$ can be rewritten as product of two quadratic polynomials in $\mathbb{Z}[X]$, then $P(x) = P_1(x)P_2(x)$. So, $P_1$ has two from the four roots of $P$. Now we divide into three cases: Case $1$: $P_1$ has $\sqrt{\alpha}$ and $-\sqrt{\alpha}$ as roots.Then, $P_1(x) = x^2 - Sx + Q$, with $Q = -\sqrt{\alpha} \times \sqrt{\alpha}$. Since $Q \in \mathbb{Z}$, then $\alpha \in \mathbb{Z} \Rightarrow n+p+2\sqrt{np} \in \mathbb{Z}$. Thus, $np$ is a perfect square and $n$ has the form $n = pk^2$. In fact, if $n = pk^2$, then $P(x) = (x^2 - p(k-1)^2)(x^2 - p(k+1)^2)$. Case $2$: $P_1$ has $\sqrt{\alpha}$ and $\pm\sqrt{\beta}$ as roots. Then, $P_1(x) = x^2 - Sx + Q$, with $S = \sqrt{\alpha} \pm \sqrt{\beta}$. So, $$S^2 = (\sqrt{\alpha} \pm \sqrt{\beta}) =\alpha + \beta \pm 2\sqrt{\alpha\beta} \Rightarrow$$$$S^2 = (2n+2p) \pm 2\sqrt{(n-p)^2} \Rightarrow$$$$S^2 = 2n + 2p \pm ( 2n - 2p)$$ If the sign is positive, then $S^2 = 4n^2$. Hence, $4n$ is a perfect square, and then $n$ has the form $n = k^2$ for some $k$. In fact, if $n = k^2$, then $P(x) = (x^2 - 2kx + k^2 - p)(x^2 + 2kx + k^2 - p)$. In the other hand, if the sign is negative, then $s^2 = 4p$, contradiction, since $p |S^2$ but $p^2 \not |S^2$. So, the solution is $n = k^2$ and $n = pk^2$ for all $k \in \mathbb{Z}$.

Roots of $P$ are $u, v, -u, -v$. Suppose $P_1 (x) = (x - u)(x + u) = x^2 - u^2$. Then $P_1 (v) = x^2 - v^2$, and so $$x^4 - 2(n + p) x^2 + (n - p)^2 = (x^2 - u^2)(x^2 - v^2).$$Therefore by discriminant, $(n + p)^2 - (n - p)^2 = 4np$ is a perfect square, which implies $n = pk^2$ for some integer $k$. Now suppose $P_1 (x) = (x - u)(x - v) = x^2 - (u+v) x + uv$. Then $P_2 (x) = (x + u)(x + v) = x^2 + (u+v) x + uv$. Compare coefficients. $$(uv)^2 = (n-p)^2 \rightarrow uv = \pm (n-p)$$$$2uv - (u+v)^2 = -2(n+p) \rightarrow (u+v)^2 = \pm 2(n-p) + 2(n+p).$$If $(u+v)^2 = -2(n-p) + 2(n+p) = 4p$ then $p$ is a perfect square, contradiction. Thus $(u+v)^2 = 2(n-p) + 2(n+p) = 4n$, so $n$ is a perfect square, $n = k^2$. Then $uv = k^2 - p, u + v = \pm 2k$. These are integers, so $P_1 (x), P_2 (x)$ are polynomials with integer coefficients. Thus $n = k^2$ or $n = pk^2$ for some integer $k$.