I'll do (b). Let $x=dx_1$ and $y=dy_1$ where $(x_1,y_1)=1$. With this, we get \[ d^2 x_1 y_1 \mid d^{2022}x_1^{2022} + dx_1 + d^2y_1^2. \]Now, we get $d^2\mid dx_1$, implying $d\mid x_1$. As $(x_1,y_1)=1$, this implies, in turn, $(d,y_1)=1$. Now using $d^2x_1\mid dx_1+d^2y_1^2 = d(x_1+dy_1^2)$, we further conclude that $dx_1\mid x_1 + dy_1^2$. As a result, $x_1\mid dy_1^2\implies x_1\mid d$. From here, we get $d=x_1$, yielding $x=d^2$.

For $(a)$ just take $x=4$ and $y=2$.

Here is the solution I submitted during the contest: a) $(x,y)=(4,2)$ works and it's easy to see that $8\mid4^{2022}+4+2^2$ since $8\mid 4^{2022}$ and $8\mid 4+2^2$. b) First of all, as we showed in a), $x$ can be a perfect square. Thus it suffices to show that $x$ can only be a perfect square. We prove by contradiction. Notation: $v_p(a)$ is the greatest non-negative integer $b$ such that $p^b\mid a$. In other words, in this case, $p^{b+1}\nmid a$. It's obvious that if $x$ is a perfect square, then $2\mid v_p(x)$ for all primes $p$. Therefore, for the sake of contradiction, assume that there exists a prime number $p$ such that $2\nmid v_p(x)$, in other words, $v_p(x)$ is odd. For convenience, we let $v_p(x)=i$ for that specific $p$. Now we have $p^i\mid x$, as $xy\mid x^{2022}+x+y^2$ it implies that $p^i\mid x^{2022}+x+y^2$, and thus $p^i\mid y^2$. Here we notice that as $y^2$ is a perfect square, $2\mid v_p(y^2)$ for any prime $p$, therefore we must have $p^{i+1}\mid y^2$. It's also obvious that $p\mid y$. So now we have $p^{i+1}\mid xy$, which yields $p^{i+1}\mid x^{2022}+x+y^2$. As obviously $p^{i+1}\mid x^{2022}$ and $p^{i+1}\mid y^2$, $p^{i+1}\mid x$, and we have a contradiction since we've already assumed that $i=v_p(x)$ which makes $p^{i+1}\nmid x$. Hence $x$ must be a perfect square. Q. E. D.

b) It's obvious that $xy|xy$ and $xy|y(x+y^2+x^{2022})$ so it must divide the difference meaning: $xy|xy+y^3+x^{2022}y-xy$ so $xy|y^3+x^{2022}y$ and dividing by y we have $x|y^2+x^{2022}$ and notice that $x|x^{2022}$ so $x|y^2$ and if we call $d=gcd (x,y)$ then $x=da$ and $y=db$ where $gcd(a,b)=1$ so $x|y^2$ $\implies$ $da|d^2b^2$ and simplifying $a|db^2$ but $gcd(a,b)=gcd(a,b^2)=1$ $\implies$ $a|d$. Now using this in the beginning we have $d^2ab|da+d^2y^2+d^{2022}a^{2022}$ and factoring $d^2$, $d^2ab|d^2(y^2+d^{2020}a^{2022}) +da$ so we notice $d^2|da$ $\implies$ $d|a$ and with $a|d$ we conclude $a=d$ so $x=d^2$.

$a)$$(x, y)=(4, 2)$ $b)$ Let $p\mid x$ be a prime. $x\mid x^{2022}+x+y^2\Rightarrow x\mid y^2\Rightarrow p\mid y\Rightarrow v_p(x)\leq 2v_p(y)$. FTSOC, assume $v_{p}(x)\neq 2v_{p}(y)$. $xy\mid x^{2022}+x+y^2\Rightarrow v_{p}(x)+v_{p}(y)\leq min\{v_p(x), 2v_p(y), v_p(x^{2022})\}=v_{p}(x)\Rightarrow v_{p}(y)\leq 0\Rightarrow Contradiction!$

Thus $v_{p}(x)=2v_{p}(y)$ $\forall$ $p\mid x\Rightarrow x$ is a square. Remark:Click to reveal hidden text

ericxyzhu wrote: Here is the solution I submitted during the contest: a) $(x,y)=(4,2)$ works and it's easy to see that $8\mid4^{2022}+4+2^2$ since $8\mid 4^{2022}$ and $8\mid 4+2^2$. b) First of all, as we showed in a), $x$ can be a perfect square. Thus it suffices to show that $x$ can only be a perfect square. We prove by contradiction. Notation: $v_p(a)$ is the greatest non-negative integer $b$ such that $p^b\mid a$. In other words, in this case, $p^{b+1}\nmid a$. It's obvious that if $x$ is a perfect square, then $2\mid v_p(x)$ for all primes $p$. Therefore, for the sake of contradiction, assume that there exists a prime number $p$ such that $2\nmid v_p(x)$, in other words, $v_p(x)$ is odd. For convenience, we let $v_p(x)=i$ for that specific $p$. Now we have $p^i\mid x$, as $xy\mid x^{2022}+x+y^2$ it implies that $p^i\mid x^{2022}+x+y^2$, and thus $p^i\mid y^2$. Here we notice that as $y^2$ is a perfect square, $2\mid v_p(y^2)$ for any prime $p$, therefore we must have $p^{i+1}\mid y^2$. It's also obvious that $p\mid y$. So now we have $p^{i+1}\mid xy$, which yields $p^{i+1}\mid x^{2022}+x+y^2$. As obviously $p^{i+1}\mid x^{2022}$ and $p^{i+1}\mid y^2$, $p^{i+1}\mid x$, and we have a contradiction since we've already assumed that $i=v_p(x)$ which makes $p^{i+1}\nmid x$. Hence $x$ must be a perfect square. Q. E. D. Thanks for explaining this clearly and in depth @ericxyzhu