**First Solution**Expanding and canceling, we have: $$x^2+y^2-2xy = x+y$$ It's easy to see that there's no solution for $x=y$. The above equation is now equivalent to: $$16x^2+16y^2+1-8x-8y+32xy = 64xy+8x+8y+1$$ $$(4x+4y-1)^2 = (8x+1)(8y+1)$$ Because both quantities are positive, we have: $$4x+4y-1 = \sqrt{8x+1} \sqrt{8y+1}$$ Once again rearranging, $$(8x+1) - 2\sqrt{8x+1} \sqrt{8y+1} + (8y+1) = 4$$ $$(\sqrt{8x+1} - \sqrt{8y+1})^2 = 4$$ If $x=y$, then there is no solution WLOG, we may assume that $x > y$, so that $$\frac{\sqrt{8x+1}}{2} - \frac{\sqrt{8y+1}}{2} = 1$$ Note that $g(t) = \frac{1+\sqrt{8t+1}}{2}$ is the reverse triangular number function. Thus, any two consecutive triangular numbers (for example 3 and 6, 6 and 10, and so on) are a solution.

Now we show that there's no other solution. The equation is equivalent to $g(x) - g(y) = 1$. In order for $g(x)$, to be an integer, we must have $8x + 1 = k^2$ for some $k$. But $k$ is odd, so $8x+1 = 4m^2 + 4m + 1$ which means $x = m(m+1)/2$ is a triangular number

Now suppose $g(x)$ is not an integer, then it will be irrational, and so will $g(y)$. They both have form of $a+\sqrt{b}$ with $a,b$ rational, but with different $b$s, because $x \neq y$. The difference can't be rational, a contradiction.

Thus, the only solutions are two consecutive triangular numbers.
**Second Solution**
Let $d = \gcd(x,y)$, and $x = dm, y = dn$ with $m,n$ coprime. Upon substitution and rearranging, we have:
$$d(m-n)^2 = m+n$$
Clearly $m=n$ is not a solution. Now wlog let $m < n$ and $n = m + k$
$$dk^2 = 2m+k$$
Because $k$ divides $dk^2$ and $2m$ then $k$ divides $2m$. Because $m$ and $n$ are coprime, then $k$ and $m$ are coprime. Thus we have $k=1$ or $k=2$.

For $k=1$, $d = 2m+1, n=m+1$, so $x = (2m+1)m,y=(2m+1)(m+1)$.

For $k=2$, $d = (m+1)/2, n=m+2$, and obviously we can only use odd values of $m$. Suppose $m = 2l+1$, then $d = l+1, n=2l+3$, so $x=(l+1)(2l+1), y =(l+1)(2l+3)$. Interestingly, this form is the same as the case for $k=1$, except we substitute $m = l + 1/2$

Combining the two cases, we can conclude that the solution must have form of $m(2m+1),(m+1)(2m+1)$ for $m$ multiples of 1/2, or equivalently, $k(k+1)/2, (k+1)(k+2)/2$ for $k$ any integers. In other words, they must be two consecutive triangular numbers.