nonlinear diophantine equation in two variables

Question

Consider the following equation in non-negative integers

$$3x^2+1=2^y$$ I suspect the only solutions are $(x,y)=(0,0),(1,2)$ but I cannot prove it.

Answer 1

If $y\ge 3$, then $\text{RHS}=2^y\equiv 0\pmod 8$.

On the other hand, since we have $x^2\equiv 0,1,4\pmod 8$, we have $$\text{LHS}=3x^2+1\not\equiv 0\pmod 8.$$

So, $y$ has to be smaller than $3$.