Okay, let's start with the polynomial ring and try to define an isomorphism to the product ring. The multiplicative identity needs to go to the multiplicative identity, for starters:
\[1 \mapsto (1,1)\]
The only remaining question is where $X$ goes. Well, we have $X^2 = 1$, so $X$ has to go to an element that squares to the identity. If it's $(1,1)$ or $(-1,-1)$ the homomorphism won't be injective, so it's either $(1,-1)$ or $(-1,1)$, and it doesn't really matter which:
\[X \mapsto (1,-1)\]
Now a general element of the polynomial ring can be written $a + bX$ for integers $a$ and $b$, and it'll be mapped to: \[(a + b, a - b)\]
But the difference between these two entries is $2b$, so we can never make any element of $\mathbb Z \times \mathbb Z$ whose entries differ by an odd number. In particular, the proposed homomorphism does not hit $(1,0)$ and is not surjective.