Krdytkyu

Here is my attempt:
$xyx neq xyxy neq xyxyx neq xyxyxy....$ gives infinite many elements in the group.
Is this correct?

You're on the right track! Let $text{FG}(x,y)=langle x,yrangle$, $G=langle x,ymid x^2,y^2rangle$, and $H$ be the minimal normal subgroup of $text{FG}(x,y)$ containing both $x^2$ and $y^2$. There is a canonical quotient homomorphism $eta:text{FG}(x,y)to G$.

We know that every element of $text{FG}(x,y)$ may be uniquely presented as a concatenation of $x,y,x^{-1},y^{1}$ such that no $x$ borders an $x^{-1}$ and likewise for $y$. For an arbitrary element $gin text{FG}(x,y)$ denote this presentation as $overline{g}$.

Consider the following algorithm $A$: given $gin text{FG}(x,y)$ we replace each instance of $x^{-1}$ and $y^{-1}$ in $overline{g}$ with $x$ and $y$ respectively. We then remove stepwise the leftmnost instances of $xx$ and $yy$ present until no such substrings remain. To take concrete examples, $$A:x^{-1}yyxxmapsto xoverbrace{yy}xxmapsto overbrace{xx}xmapsto x$$ $$A:x^{-1}yyxxxmapsto xoverbrace{yy}xxxmapsto overbrace{xx}xxmapsto overbrace{xx}mapsto text{empty}$$ Where "$text{empty}$" is the empty string.

Denote $H'$ the subset of elements $text{FG}(x,y)$ for whom $A$ outputs the null string. We can show that $H'$ is a subgroup of $text{FG}(x,y)$ and in particular a normal subgroup of $text{FG}(x,y)$. Let $eta':text{FG}(x,y)to text{FG}(x,y)/H'$ be the associated quotient map. Then we have that $eta'$ factors through a morphism $kappa:Gto text{FG}(x,y)/H'$ in the sense that $eta'=kappacirc eta$. Our $kappa$ is therefore surjective (since $eta'$ is surjective). So it suffices to demonstrate that $text{FG}(x,y)/H'$ is infinite.

To that end consider the sequence of elements $$eta'(xyx) neq eta(xyxy)' neq eta'(xyxyx) neq eta'(xyxyxy)dots$$
We will prove that the elements of this sequence are pairwise distinct. Let $eta'(g_1)$ and $eta'(g_2)$ be two such elements. It suffices to show that $eta'(g_1)eta'(g_2)^{-1}neq 1iff A(g_1g_2^{-1})neq text{empty}$. Can you finish from here?

It's worth noting that $H=H'$ and thus that $Gcong text{FG}(x,y)/H'$. The proof isn't that hard. (Specifically we've proven that $Hsubseteq H'$, proving that $H'subseteq H$ would do the trick.)