Diophantine Equation: $4x^r = 3y^2 + 1$

Question

If $r \ge 3$ is an integer, show that $4x^r = 3y^2 + 1$ does not have positive integer solutions $(x, y)$ except for $(1, 1)$. (I am not sure whether this is an open problem; in any case, it is a slightly stronger but probably more familiar form of this question.)