No. of solutions - 4

$y^{2}=1+x+x^{2}$

$x^{2}+x+(1-y^{2})=0$ .

Let this be a quadratic equation related to $x$ .From quadratic formula we have :

$x=\frac{-1 \pm \sqrt{1-4*1*(1-y^{2})}}{2}$ .

$x=\frac{-1 \pm \sqrt{4y^{2}-3}}{2}$ .

Since $x$ must be an integer then the discriminant must be a perfect square.We have the equation:

$4y^{2}-3=m^{2}$ for some integer $m$ .

$(2y+m)(2y-m)=3$ .

Since both factors are integers we have the following equations:

$2y+m=3$

$2y-m=1$ .

Solving these equations we get $y=1$ .Plugging these value into the quadratic formula we get $x=0,-1$ .Note that if in the above system we change the places of the constants ie $3$ and $1$ ,this wont affect the problem at all producing in the end the same solutions.Lets proceed with our solution. There is also another system of equations we need to consider since the factors are integers and not $positive$ integers.That is:

$2y+m=-3$

$2y-m=-1$ .

Solving we get $y=-1$ .Plugging in this value into our quadratic formula we get $x=0,-1$ . Thus ,there are in total $\boxed{4}$ solutions which are :

$(0;1),(-1;1),(-1;-1),(0;-1)$ .

$y^2=1+x+x^2$ can be rearranged into $y^2-1=x+x^2$ which can be written as $(y+1)(y-1)=x(x+1)$ . From this we can see that if $x=0, y= -1$ and $x=0, y=1$ . Also, if $x=-1, y=1$ and $x=-1, y=-1$ . These are already four solutions so there is no need to check for more since that is the highest option. So the solution is $\boxed{4}$

Isn't the answer hidden in the question name?