Multiplying both sides by two, we have $8x^3-6x+2=4y^2=(2x)^3-3(2x)+2=(2x-1)^2(2x+2)=(2y)^2$. For this to have integer solutions, we must have that $2x+2$ be a perfect square. The least positive integer solution for $x$ is $x=1$, where $2x+2$ is a square. We can see that each $x$ (let us denote the $n$-th integer solution of $x$ as $a_n$) can be expressed as $a_n=2n^2-1, a_1=1$. Since $x \leq 1980$, $2n^2-1 \leq 1980$. Solving for the maximum of $n$, we have that $\max n \approx 31.4722$, which means that there are at least $31$ possible integer values of $x$ that yield an integer $y$. To prove the variant, we note that $2n^2-1$ is monotonically increasing for positive $n$, thus there are an infinite number of integer $x$ that yield an integer $y$.