I have another way more logical than Phan Thanh Tinh's answer ! :)

Here it is : 

\(x^2+2005x+2006y^2+y=xy+2006xy^2+2007\)

\(\Leftrightarrow\left(x^2+2005x-2006\right)+\left(2006y^2-2006xy^2\right)+\left(y-xy\right)=1\)

\(\Leftrightarrow\left(x-1\right)\left(x+2006\right)-2006y^2\left(x-1\right)-y\left(x-1\right)=1\)

\(\Leftrightarrow\left(x-1\right)\left(x+2006-2006y^2-y\right)=1\) (1)

Cause x,y are interger numbers so 2 factors on the left side of expression (1) is divisor of 1.

So it occurs two cases :

Case 1 : \(\left\{{}\begin{matrix}x-1=1\left(2\right)\\x+2006-2006y^2-y=1\left(3\right)\end{matrix}\right.\)

From (2) -> x = 2.

(3) \(\Leftrightarrow2006y^2+y-2007=0\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-\dfrac{2007}{2006}\notin Z\end{matrix}\right.\)

So (x;y) = (2;1).

Case 2 : \(\left\{{}\begin{matrix}x-1=-1\left(4\right)\\x+2006-2006y^2-y=-1\left(5\right)\end{matrix}\right.\)

From (4) => x = 0.

(5) \(\Leftrightarrow2006y^2+y-2007=0\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-\dfrac{2007}{2006}\notin Z\end{matrix}\right.\)

So (x;y) = (0;1).

So the expression has two pairs of interger numbers (x;y) are (2;1) and (0;1).

\(x^2+2005x+2006y^2+y=xy+2006xy^2+2007\)

\(\Leftrightarrow xy+2006xy^2+2007-x^2-2005x-2006y^2-y=0\)

\(\Leftrightarrow\left(xy-y\right)+\left(2006xy^2-2006y^2\right)-x^2+2x-1-2007x+2007=-1\)

\(\Leftrightarrow y\left(x-1\right)+2006y^2\left(x-1\right)-\left(x-1\right)^2-2007\left(x-1\right)=-1\)

\(\Leftrightarrow\left(x-1\right)\left[y\left(2006y+1\right)-x-2006\right]=-1\)

Case 1 : \(x-1=1\Rightarrow x=2\) and :

\(y\left(2006y+1\right)-2-2006=-1\Leftrightarrow y\left(2006y+1\right)=2007\)

2006y + 1 is a divisor of 2007,so \(-2007\le2006y+1\le2007\)

\(\Rightarrow-1\le y\le1\).Since \(y\ne0\),we have :

\(\left[{}\begin{matrix}y=-1\\y=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2006y+1=-2005\\2006y+1=2007\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}y\left(2006y+1\right)=2005\\y\left(2006y+1\right)=2007\end{matrix}\right.\)

\(\Rightarrow y=1\)

Case 2 : \(x-1=-1\Rightarrow x=0\) and :

\(y\left(2006y+1\right)-2006=1\Rightarrow y\left(2006y+1\right)=2007\Rightarrow y=1\)

Hence,\(\left(x;y\right)=\left(0;1\right);\left(2;1\right)\)