$\displaystyle \displaystyle \begin{align*} x^2 - 4x\,y + 5y^2 + 2y - 4 &= 0 \\ x^2 - 4x\,y + \left(-2y\right)^2 - \left(-2y\right)^2 + 5y^2 + 2y - 4 &= 0 \\ \left(x - 2y\right)^2 - 4y^2 + 5y^2 + 2y - 4 &= 0 \\ \left(x - 2y\right)^2 + y^2 + 2y - 4 &= 0 \\ \left(x - 2y\right)^2 + y^2 + 2y + 1^2 - 1^2 - 4 &= 0 \\ \left(x - 2y\right)^2 + \left(y + 1\right)^2 - 5 &= 0 \end{align*}$
This is an ellipse. It's clear there are an infinite number of points that satisfy this equation, since the relation is continuous.
(x - 2y)^2 + (y + 1)^2 - 5=0 - Wolfram|Alpha
thanks to Prove It for all the grunt work ...
the original question (in the attachment) asks for total number of integer ordered pairs (x,y)
so, two possible cases to consider ...
$\displaystyle (x-2y)^2 = 4 \, , \, (y+1)^2 = 1$
or
$\displaystyle (x-2y)^2 = 1 \, , \, (y+1)^2 = 4$
... leave it for you to finish.