Let $p\in k[x_1, \dots, x_n]$ such that p does not involve the variable $x_i$ and is not square of any polynomial. Then we want to show that the polynomial $f = x_i^2-p$ is irreducible.
I tried assuming otherwise, assuming $f=gh$ for some non-constant polynomials $g,h\in k[x_1, \dots, x_n]$. Then differentiating both sides with respect to $x_i$, we get
\begin{equation} \frac{\partial f}{\partial x_i} = \frac{\partial (x_i^2-p)}{\partial x_i} = 2x_i = \frac{\partial g}{\partial x_i}h +g\frac{\partial h}{\partial x_i}
\end{equation}
This should somehow lead to a contradiction with the fact that $p$ is not square of any polynomial but I couldn't get further from here on.
Best Answer
Down to earth argument. Assume that $f=gh$. View $f,g,h$ as elements of $R[x_i]$, where $R=k[x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_n]$.
Then $\deg(f)=2=\deg(f)+\deg(h)$, so $\deg(g)=0,1,2$ and $\deg(h)=2,1,0$.
Note that comparing leading terms shows that the leading terms of $g,h$ are in $R^\times=k$, hence we can assume wlog that $g,h$ are monic.
In this case, if $g$ or $h$ is constant, it is equal to $1$, hence invertible, and we are done.
So we may asume that $g,h$ are monic of degree $1$. Write $g=x_i-a,h=x_i-b, a,b\in R$.
Then $f=x_i^2-p=gh=x_i^2-(a+b)x_i+ab$, so $b=-a$, and $-p=ab=-a^2$, so $p=a^2$, contradiction.
Commutative algebra argument. $R=k[x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_n]$ is a UFD, and $f$ is monic in $x_i$, so $f$ is irreducible over $R$ if and only if it is irreducible over $K=Frac(R)$.
Since $f$ has degree $2$, $f$ is irreducible over $K$ if and only if it has a root in $K$, that is, if and only if $p$ is a square in $K$.
So we have an equality $pv^2=u^2$, where $u,v\in R$ are coprime. If $v$ is not a unit of $R$, it is divisible by an irreducible element $\pi$, but then $\pi$ divides $u^2$, so it divides $u$, contradiction. Hence $v$ is a unit, and $p$ is a square in $R$, contradiction again.