$\underline{\textbf{Solution 1 (Squeeze Method)}}$

It is clearly that $x > y$. Thus, let $x-y=d$ where $d$ is a positive integer. Then the equation is equivalent to $$3y^2d + 3yd^2 + d^3 = y^2+dy+61$$ or $$(3d-1)y^2 +(3d^2-1)y+d^3=61$$ Because $y$ is positive, this relation implies $$d^3 < 61\implies d=1, 2, 3$$ When $d=1$, we have $2y^2 + 2y+1=6 \implies (x,y)=\boxed{(6, 5)}$. When $d=2$ or $3$, there is not integer solutions.

$\underline{\textbf{Solution 2 (Factorization Method)}}$ (Credit goes to Mr. Mark)

The given equation is equivalent to $$(3x-3y-1)(9x^2 + 9y^2 +3x - 3y+9xy+1)=1646=2\times 823$$ Because both $2$ and $823$ are prime and $(9x^2 + 9y^2 +3x - 3y+9xy+1) > (3x-3y-1)$, it must hold that $$ \left\{ \begin{array}{rcl} 3x-3y-1 &=& 2\\ 9x^2 + 9y^2 +3x - 3y+9xy+1&=&823 \end{array} \right. $$ Solving this system will yield $(x, y)=\boxed{(6,5)}$. Note: the second method will still work if $x$ and $y$ are not necessarily positive.