Some years ago...

As $2011$ is prime so $x-y=1$ and $x+y=2011\Rightarrow 2x=2012 \therefore x=1006$ consequently $y=1005$ .

But as we are talking of square numerbs $x=\pm 1006$ and $y=\pm 1005$ . So there are $\boxed{4}$ solutions.

$(1006,1005),(-1006,1005),(1006,-1005),(-1006,-1005)$

Write a solution. We firstly proceed with the difference of two squares identity. x^2 - y^2 = (x - y)(x +y) After this, we must notice that 2011 is prime, so we need two integers that add to make 2011, and subtract to make 1. Now, we obtain the first set of solutions (1006,1005). After playing around a little bit with the signs (due to the negative nature of the first bracket), we find 3 others. Thus, the solution is 4. ((1006,1005),(-1006,1005),(1006,-1005),(-1006,-1005))

Factor the LHS (left hand side) and introduce new variables to simplify the problem: $\begin{aligned} 2011&=x^2-y^2=(x-y)(x+y)=:uv&&&&\left|\begin{pmatrix}u\\ v\end{pmatrix}:=\begin{pmatrix} 1&-1\\ 1&1 \end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}\in\mathbb{Z}^2\right. \end{aligned}$ Notice 2011 is prime and only has two positive factors 1 and 2011. Sign combinations lead to a total of $\boxed{4}$ solutions: $\begin{aligned} u&\in\{\pm1;\:\pm2011\},&v&=\frac{2011}{u}:&&&\begin{pmatrix}x\\ y\end{pmatrix}&=\frac{1}{2}\begin{pmatrix} 1&1\\ -1&1 \end{pmatrix}\begin{pmatrix}u\\ v\end{pmatrix}\in\left\{ \begin{pmatrix}\pm1006\\ \pm1005\end{pmatrix} \right\} \end{aligned}$