smc

Let $a=x^2$ and $b=y^2$. We now have a system of 2 linear equations: $$\begin{gathered} a+4b=1 \\ 4a+b=4 \end{gathered}$$ Multiplying the first equation by 4 and then subtracting the second equation from the first one, we get: $$\begin{gathered} 15b=0 \\ b=0 \end{gathered}$$ Now, we can substitute b to solve for a: $$\begin{gathered} 4a+0=4 \\ a=1 \end{gathered}$$ Now note that $x^2=1\to x=\pm 1$ and $y^2=0\to y=0$, so the solutions are $(1,0),(-1,0)$. There are two solutions, meaning that the answer is $\boxed{2}$