Let us work in the Gaussian integers $\mathbb{Z}[i]$. Firstly note that the roots of $P(x)+1$ are $S=\{k(4\pm 3i)|1\leq k\leq n\}=\{s_1,...,s_{2n}\}$, which belong to $\mathbb{Z}[i]$. Now assuming $P(x)=A(x)B(x)$, with $A,B\in \mathbb{Z}[x]$, we must have $A(s)B(s)=1$ for every $s\in S$. This means that either $A(s)=B(s)=\pm 1$ or $A(s)=-B(s)=\pm i$, and so assume that $S$ can be partitioned in $U=\{s\in S| A(s),B(s)\in\{\pm 1\}\}=\{u_1,..,u_M.\}$ and $V=\{s\in S| A(s),B(s)\in\{\pm i\}\}=\{v_1,...,v_N\}$. Since $A(u_i)=B(u_i)$ it follows that $A(x)-B(x)=k(x)(x-u_1)\cdots(x-u_M)$ for some $k\in \mathbb{Z}[i][x]$. Similarly $A(x)+B(x)=l(x)(x-v_1)\cdots(x-v_N)$ for some $l\in \mathbb{Z}[i][x]$. Therefore we can get $A(x)=\frac{k(x)(x-u_1)\cdots(x-u_M)+l(x)(x-v_1)\cdots(x-v_N)}{2}$ and if we plug one of the $u_i$ we must get $1=|A(u_i)|=|\frac{l(u_i)(u_i-v_1)\cdots (u_i-v_N)}{2}|$. However we note that for distinct $x,y\in S$ we have $|x-y|\geq |3+4i|=5$, so the product is at least $\frac{5}{2}>1$, which is a contradiction. The only case left is when $U$ or $V$ are empty. Taking the first case (the other is the same thing) , we would have $deg A+B\geq 2n$ (because $(x-s_1)\cdots(x-s_{2n})|A+B$)but this implies $deg A=2n$ or $degB=2n$ which is a contradiction since the other factor would be constant. Therefore $P$ is irreducible in $\mathbb{Z}[x]$