It's clear by inspection that $x$ and $y$ must have the same parity, so let's write $y=x+2k$. The equation reduces to a quadratic in $x$:
The discriminant of this quadratic equation is
For the quadratic to have any real (much less integer) roots, $\Delta(k)$ must be non-negative. This clearly restricts $k$ to a finite interval, which turns out to be $1\le k\le5$. (It's easy to see $\Delta(k)\lt0$ if $k\le0$. The verification that $\Delta(k)\lt0$ if $k\ge6$ is little messy, so I'm omitting it. However, you can, if you like, skip from here to the "added later" section below.)
For the quadratic to have any integers roots, $\Delta(k)$ must be a square, so at this point it's easiest to just calculate:
The only square here is when $k=1$, so we see the quadratic
which has roots $x=0$ and $x=9$. Thus the only solution in positive integers is $(x,y)=(9,11)$, and the only other solution in integers is $(x,y)=(0,2)$.
Remark: This solution and mathlove's are quite different, but they both lead to computing a small number of cases (nine in mathlove's, five here).
Added later: I got to wondering if there's some easy way to avoid doing a case-by-case computational check. It turns out there is, at least if you restrict to $x\gt0$ (which is specified by the OP).
Starting from the observation that $k$ must be positive in order for $\Delta(k)$ to be non-negative, let's go back to the quadratic in $x$ and rewrite it as a cubic in $k$:
Thinking of the left hand side as a function $f(k)$ (holding $x$ constant), note that
Consequently, if $k\ge2$,
This leaves $k=1$ as the only possible (integer) value that allows for a positive integer value of $x$, which turns out to be $x=9$ (hence $y=11$).