Take $f=(2t^2-1)^2, g=2(2t^2-1)^2$. There are infinitely many odd primes $p$ such that 2 is not a square mod $p$. Hence $f(x)\equiv g(y) (p)\iff (2x^2-1)^2 \equiv 2(2y^2-1)^2 (p)\implies 2x^2-1\equiv 2y^2-1\equiv 0(p)\implies 2x^2\equiv 2y^2\equiv 1(p)$, contradiction.

An even simpler construction is $f(x)=x^2$ and $g(y)=-(y^2+1)^2$. Then, if for a prime $p \equiv 3 \pmod 4$ we have $p \mid x^2+(y^2+1)^2,$ then we must have $p \mid x$ and $p \mid y^2+1$, and subsequently $p \mid 1,$ a contradiction.

Orestis_Lignos wrote: An even simpler construction is $f(x)=x^2$ and $g(y)=-(y^2+1)^2$. Then, if for a prime $p \equiv 3 \pmod 4$ we have $p \mid x^2+(y^2+1)^2,$ then we must have $p \mid x$ and $p \mid y^2+1$, and subsequently $p \mid 1,$ a contradiction. how did you get p=1?

If $p\equiv 3 \pmod{4}$ and $p|x^2+y^2$, then $p|x$ and $p|y$. ($p|x^2+1^2\rightarrow p|1$)