Answer: D

Small Hint:

Rewrite $P=(x^2+3x+1)^2-3(x-10)$

Big Hint:

Distinct integer squares $U^2$ and $V^2$ differ by at least $2\max(|U|,|V|)-1$

Solution:

Let $N=x^2+3x+1.$ Then $$P=N^2-3(x-10).$$ At $x=10,\ P=N^2=131^2,$ so one value works.

If $x\gt10$ and $P=y^2,$ then $N^2-y^2=3(x-10).$ Distinct integer squares differing from $N^2$ differ by at least $2|N|-1,$ which is already greater than $3(x-10),$ a contradiction. If $x\lt10,$ the analogous bound $$\begin{gathered} 3(10-x)=y^2-N^2\\ \geq2|y|-1\gt2|N|-1 \end{gathered}$$ can hold only for $-6\leq x\leq2.$ Direct substitution for these nine integers gives no square. Hence $x=10$ is the unique solution and $n=1.$

Therefore, the correct answer is D.