$[(x-a_1)(x-a_2) \cdots (x-a_n)]^2 +1$ is irreducible over $\mathbb Q$

This is an old (1909) result of Issai Schur which can be found in Pólya-Szegő , on page 133 of Volume 1 . Here is the proof of irreducibility:

Let $f(x)=(x-a_1)^2(x-a_2)^2 \cdots (x-a_n)^2 +1$ be your polynomial and suppose it factors non-trivially as $f(x)=g(x)h(x)$ over $\mathbb Q$.
By Gauss's lemma we may assume that $g,h$ are monic with integral coefficients: $$g(x)=x^k+b_{k-1}x^{k - 1}+\cdots+b_0,\;h(x)=x^l+c_{l-1}x^{l-1}+\cdots+c_0\in \mathbb Z[x] $$ Notice that the polynomial functions functions $g,h$ satisfy $g(r),h(r)\gt 0$ for $r\in \mathbb R$ and that $g(a_i)=h(a_i)=1$ .
We may assume $k\leq l $, so that $k\leq n$ since $k+l=2n$, and then we distinguish two cases:

Case 1: $k\lt l$
Then the polynomial $g$ takes the value $1$ for the $n$ distinct values $a_1,\cdots a_n$ and, because it has degree $k\lt n$, that polynomial is the constant $g=1$ and the factorization $f=gh$ is trivial, contrary to our assumption .

Case 2: $k=l=n$
Then $g-h$ is a polynomial of degree $\lt n$ vanishing at the $n$ numbers $a_i$, so that $g-h=0$ and $g=h$.
Hence the assumed factorization $f=gh$ becomes $f(x)=(x-a_1)^2(x-a_2)^2 \cdots (x-a_n)^2+1 =g(x)^2$ and we get $ 1= g(x)^2-(x-a_1)^2(x-a_2)^2 \cdots (x-a_n)^2$ so that $$1=[g(x)+(x-a_1)(x-a_2) \cdots (x-a_n)][g(x)-(x-a_1)(x-a_2) \cdots (x-a_n)]$$ This is a clearly absurd factorization of $1$ into positive degree polynomials.

Conclusion
In both cases the supposed non-trivial factorizability of $f$ leads to a contradiction and thus $f$ is actually irreducible.