Check Proof: (2,x,y) is a prime ideal in $\mathbb{Z}[x,y]$ (2,x,y) is a prime ideal in $\mathbb{Z}[x,y]$
Assume $a\in \mathbb{Z}[x,y]$, $b \in \mathbb{Z}[x,y]$ and $ab \in (2,x,y)$
Assume $a \notin (2,x,y)$ and $b \notin (2,x,y)$
Then, both $a$ and $b$ has a term which does not have $x$, $y$ or a constant which is a multiple of 2. 
Then, $ab$ has a term which does not have either $x$, $y$ or a constant which is a multiple of 2. 
Contradiction as $ab \in (2,x,y)$. 
 A: By the Lemma: $\ fg\in I \color{#c00}\iff 2\mid f_0 g_0\iff 2\mid f_0\ \ {\rm or}\ \ 2\mid g_0\color{#c00}\iff f\in I\ \ {\rm or}\ \ g\in I$
Lemma $\ \ f \in I=(2,x,y) \color{#c00}\iff 2\mid f_0 := f(0,0).\ \ $ Proof: $ $ Reducing modulo $I$ we have 
$\!\!\bmod I\!:\,\ x\equiv0\equiv y\,\Rightarrow\, f(x,y)\equiv f(0,0)=f_0,\,$ so $\,f\in I\!\iff f_0 \in I\iff 2\mid f_0,\, $ since
$f_0 \in I \,\Rightarrow\, f_0 = 2g\! +\! xh\! +\! y k\,\Rightarrow\, f_0 = 2g_0\,$ by eval at $\,x=0=y;\,$ conversely $\, f_0 = 2n\in I$
Remark $\ $ If you know quotient rings then the above elementwise proof can be reformulated structurally as: $\ \Bbb Z[x,y]/(2,x,y)\cong \Bbb Z_2\,$ is a domain so $\,(2,x,y)\,$ is prime. The ring isomorphism can be proved using the standard isomorphism theorems.
A: You might want to clarify how you get from the observation that

Then, both $a$ and $b$ has a term which does not have $x$, $y$ or a constant which is a multiple of $2$.

to the conclusion that

Then, $ab$ has a term which does not have either $x$, $y$ or a constant which is a multiple of $2$.

You could give a more explicit description of what such elements of $\Bbb{Z}[x,y]$ look like.
