The key to the problem is to consider Modulo 3 and the fact that $3 \mid P(x+3)-P(x)$ Case 1: $u \equiv 0 \pmod{3}$ We get \begin{align*} P(P(u)) &\equiv P(P(-3)) \pmod{3} \\ u^2 &\equiv P(-40) \pmod{3} \\ u^2 &\equiv P(-1) \pmod{3} \\ u^2 &\equiv 2 \pmod{3} \end{align*}Contradiction Case 2: $u \equiv 1 \pmod{3}$ We get \begin{align*} P(P(u)) &\equiv P(P(-5)) \pmod{3} \\ u^2 &\equiv P(-156) \pmod{3} \\ u^2 &\equiv P(-3) \pmod{3} \\ u^2 &\equiv 2 \pmod{3} \end{align*}Contradiction. Case 3: $u \equiv 2 \pmod{3}$ We get \begin{align*} P(P(u)) &\equiv P(P(-1)) \pmod{3} \\ u^2 &\equiv P(-4) \pmod{3} \\ u^2 &\equiv P(-1) \pmod{3} \\ u^2 &\equiv 2 \pmod{3} \end{align*}Contradiction. Therefore, the largest possible number of integers $x$ satisfying P(P(x)) = x^2$ is $0$.

Gems98 wrote: The key to the problem is to consider Modulo 3 and the fact that $3 \mid P(x+3)-P(x)$ Case 1: $u \equiv 0 \pmod{3}$ We get \begin{align*} P(P(u)) &\equiv P(P(-3)) \pmod{3} \\ u^2 &\equiv P(-40) \pmod{3} \\ u^2 &\equiv P(-1) \pmod{3} \\ u^2 &\equiv 2 \pmod{3} \end{align*}Contradiction Case 2: $u \equiv 1 \pmod{3}$ We get \begin{align*} P(P(u)) &\equiv P(P(-5)) \pmod{3} \\ u^2 &\equiv P(-156) \pmod{3} \\ u^2 &\equiv P(-3) \pmod{3} \\ u^2 &\equiv 2 \pmod{3} \end{align*}Contradiction. Case 3: $u \equiv 2 \pmod{3}$ We get \begin{align*} P(P(u)) &\equiv P(P(-1)) \pmod{3} \\ u^2 &\equiv P(-4) \pmod{3} \\ u^2 &\equiv P(-1) \pmod{3} \\ u^2 &\equiv 2 \pmod{3} \end{align*}Contradiction. Therefore, the largest possible number of integers $x$ satisfying P(P(x)) = x^2$ is $0$. Nice solution , but dont you think that even x=0 is not possible ?