Is the polynomial determinant of linear independent polynomials irreducible? Let $m,n\in\mathbb N $, and let  $A=[p_{ij}]_{i,j=1}^m$ be a matrix of irreducible polynomials where $p_{ij} \in \mathbb{R}[x_1, \dots, x_n]$. 
Depending on $m$ and $n$, prove or disprove that if $p_{ij}$ are linearly independent polynomials then the determinant polynomial $|A|$ is irreducible over reals.
Thanks in advance.   

For $m = 2$, $n = 1$, there is a counterexample.
$$  \det \begin{pmatrix} x^2 + 1 & x^2 + 2 \\ x^2 + 3 & x^2 + 5 \end{pmatrix}  = x^2 - 1 = (x+1)(x-1)  \text{.}  $$
This is also a counterexample for the problem as stated for $n > 1$ (since there is no requirement that all $n$ variables be used).
For $m= 3$, $n>1$, there are counterexamples.
$$  \det \begin{pmatrix} x^2 + 1 & x^2 + 2 & x^2 + 4 \\ x^2 + 3 & x^2 + 5 & x^2 + 7 \\ x^2 + 6 & x^2 + 9 & x^2 +10  \end{pmatrix}  = x^2 - 1 = (x+1)(x-1)  \text{.}  $$
$$  \det \begin{pmatrix} x^2 + 1 & x^2 + 2 & x^2 + 4 \\ x^2 + 3 & x^2 + 5 & x^2 + 7 \\ x^2 + 6 & x^2 + 9 & x^2 +12  \end{pmatrix}  = x^2 - 3 = (x+\sqrt{3})(x-\sqrt{3})  \text{.}  $$
The claim seems somewhat dubious.  (These were not found by any sort of search, they were the second or third thing I tried for each $m$.)
