If $x+y^3,x^2+y^2,x^3+y$ are all integers, are $x,y$ both integers?

Here's a hint (or at least some useful steps that are too long for a comment): Let $t=xy$. Note that $$\mathbb Z\ni (x^3+y)(x+y^3)-(x^2+y^2)^2=t^3+t-2t^2=t(t-1)^2$$ and \begin{align*} \mathbb Z &\ni \big[(x^3+y)(x^2+y^2)-(x+y^3)\big]\big[(x+y^3)(x^2+y^2)-(x^3+y)\big]\\ &=t(t-1)^2+(x^2+y^2)^2t(t^2+t-1). \end{align*} Therefore $t^3+t^2-t$ is rational (since $x^2+y^2\neq 0$ or we are done). Can you finish from here?

In this solution $x$ and $y$ are not assumed to be real numbers. I will find all $(x,y)\in\Bbb C^2\setminus\Bbb Z^2$ such that $x^3+y$, $x+y^3$, and $x^2+y^2$ are integers. The claim is that all possible $(x,y)$ are given below by $(3)$ with $|d|\ge2$, and $(4)$ for any integer $d$, and none of such solutions satisfy $(x,y)\in\mathbb{R}^2$.

Let $a=x^3+y$, $b=x+y^3$, and $c=x^2+y^2$. Observe that $x$ is a root of $$A(t)=(t^3-a)^3-t+b$$ and $y$ is a root of $$B(t)=(t^3-b)^3-t+a.$$ Therefore both $x$ and $y$ are algebraic integers.

If $x=0$, or $y=0$, then it is easy to see that $x$ and $y$ are integers. If $x= \pm y$, then $x^2=\frac{c}{2}$, so $$a=x^3+y=\frac{c}{2}x\pm x=\frac{c\pm 2}{2}x.$$ If $c=2$, then $x^2=\frac{c}{2}=1$, so $x$ is an integer, making $y$ also an integer. If $c=-2$, then $x^2=\frac{c}{2}=-1$ so $x=\pm i$. Therefore we have $2$ solutions $$(x,y)=\pm(i, i).\tag{1}$$ If $c\ne \pm2$, then $x$ is rational. A rational algebraic integer is necessarily an integer. Thus, $x$ and $y$ are both integers.

Suppose now that $x\ne \pm y$, $x\ne 0$, and $y\ne 0$. If $a=b$, then $$(x-y)(x^2+xy+y^2-1)=(x^3+y)-(x+y^3)=a-b=0.$$ Since $x\ne \pm y$, we get $x^2+xy+y^2=1$, so $$xy=1-x^2-y^2=1-c.$$ Hence $x^2$ and $y^2$ are roots of the quadratic $$Q(t)=t^2-cx+(1-c)^2.$$ Note that $$x^3+y=x(x^2+y^2)-xy^2+y=cx-(1-c)y+y.$$ So $$a=c(x+y).$$ If $c=0$, then $a=0$ so $b=0$. Now, $Q(t)=t^2+1$, so $x^2,y^2=\pm i$. Therefore $$x,y=\pm\frac{1}{\sqrt2}\pm\frac{i}{\sqrt2}.$$ Because $x^2+y^2=c=0$, we get $4$ solutions: $$(x,y)=\left(e^{i\frac{k\pi}{4}},e^{-i\frac{k\pi}{4}}\right),\tag{2}$$ where $k=\pm1,\pm3$. If $c\ne 0$, then $x+y=\frac{a}{c}$ and $xy=1-c$. Hence the algebraic integers $x$ and $y$ are roots of $$P(t)=t^2-\frac{a}{c}t+(1-c).$$ Note that this means $\frac{a}{c}$ is an integer. We write $a=cd$. Then $$c=x^2+y^2=(x+y)^2-2xy=d^2-2(1-c)$$ so $$d^2=c+2(1-c)=2-c.$$ Hence $$P(t)=t^2-dt+(1-c)=t^2-dt+(d^2-1).$$ The roots are $$x,y=\frac{d\pm\sqrt{4-3d^2}}{2}.$$ Thus there are $2$ solutions $$(x,y)=\left(\frac{d\pm\sqrt{4-3d^2}}{2},\frac{d\mp\sqrt{4-3d^2}}{2}\right).\tag{3}$$ Note that if $x$ and $y$ are both real, then $d=0$ or $d=\pm1$, but this means $x$ and $y$ are integers. Non-integer solution given by $(3)$ must come from $|d|\ge 2$. Solutions $(2)$ can be included in $(3)$ by taking $d=\pm 2$.

If $a=-b$, then $$(x+y)(x^2-xy+y^2+1)=(x^3+y)+(x+y^3)=a+b=0.$$ Because $x\ne \pm y$, we get $x^2-xy+y^2=-1$ so $$xy=x^2+y^2+1=c+1.$$ Hence $x^2$ and $y^2$ are roots of the quadratic $$Q(t)=t^2-cx+(c+1)^2.$$ Note that $$x^3+y=x(x^2+y^2)-xy^2+y=cx-(c+1)y+y.$$ So $$a=c(x-y).$$ If $c=0$, then $a=0$ so $b=0$, and we are back to solutions $(2)$. If $c\ne 0$, then $x-y=\frac{a}{c}$ and $x(-y)=-(c+1)$. Hence the algebraic integers $x$ and $-y$ are roots of $$P(t)=t^2-\frac{a}{c}t-(c+1).$$ Note that this means $\frac{a}{c}$ is an integer. We write $a=cd$. Then $$c=x^2+y^2=(x-y)^2+2xy=d^2+2(c+1)$$ so $$d^2=c-2(c+1)=-2-c.$$ Hence $$P(t)=t^2-dt-(c+1)=t^2-dt+(d^2+1).$$ The roots are $$x,-y=\frac{d\pm\sqrt{-4-3d^2}}{2}.$$ Thus there are $2$ solutions $$(x,y)=\left(\frac{d\pm\sqrt{-4-3d^2}}{2},-\frac{d\mp\sqrt{-4-3d^2}}{2}\right).\tag{4}$$ None of the solutions given by $(4)$ are real. Solutions $(1)$ can be included in $(4)$ by taking $d=0$.

Let now suppose that $x\ne \pm y$, $x\ne 0$, $y\ne 0$, and $a\ne \pm b$. $$a=x^3+y=(x^2+y^2)x+(1-xy)y=cx+(1-p)y$$ and $$b=x+y^3=(1-xy)x+(x^2+y^2)y=(1-p)x+cy.$$ Since $a\ne \pm b$, $p\ne 1\pm c$. If $p$ is an integer and $p\ne 1\pm c$, then the system above yields rational solutions $(x,y)$, which means that $x,y$ are integers. We aim to show that $p$ is an integer.

If $c=0$, then $x^2+y^2=0$ so $y=\pm xi$. We have $$a=x^3+y=x^3\pm xi$$ and $$b=x+y^3=x\mp x^3i=\mp (x^3\pm xi)i=\mp ai.$$ This can happen only when $a=b=0$, but this brings us back to solutions $(1)$. We now suppose that $c\ne0$.

The rest uses the other answer by Carl Schildkraut. If $p=xy$, then we have that $p=xy$ is a root of the polynomials $$S(t)=t(t-1)^2-k$$ and $$T(t)=t^3+t^2-t-r$$ for some $k\in\Bbb Z$ and $r\in\Bbb Q$. Thus $p$ is a root of $$M(t)=\frac{T(t)-S(t)}{3}= t^2-\frac23t+\frac{k-r}{3}.$$ Note that $p$ is an algebraic integer because it is a root of $S(t)$. Therefore $p$ is a root of a monic irreducible polynomial $R(t)\in \Bbb Z[t]$. We must have $R(t)\mid M(t)$. Since $M(t)\notin\Bbb{Z}[t]$, $R(t)\ne M(t)$, so $R(t)$ must be linear. That is, $p$ is an integer.