show this equation $3x^4-y^2=3$ has no integer solution show this diophantine equation
$$3x^4-y^2=3$$ has no integer $(y\neq 0)$ solution?
My try: WLOG Assmue $(x,y)$ is postive integer  solution,then $3|y$,let $y=3y'$,then we have
$$x^4-1=3y'^2\tag{1}$$
and following I want $\pmod5$,since $x^4\equiv 0,1\pmod 5$
(1):if $x^4\equiv 0\pmod 5$or $5|x$,then $(1)$ it is clear no solution
(2)if $(x,5)=1$,then $x^4-1\equiv 0\pmod 5$,there also exsit $y'$ such $5|3y'^2$,so How to solve my problem? Thanks

I tried the following.
$3x^4-y^2=3$
$3x^4-3=y^2$
$3(x^2-1)(x^2+1)=y^2 \tag{1}$
If $y^2$ is divisible by $3$ then one of the factors of $(x^2-1)(x^2+1)$ should be divided by $3$ to make the right hand side a perfect square. So $x$ can't be $0 \pmod{3}$. Now if we divide 3 from the both hand side we can see the following.
$$ (x^2-1)(x^2+1)=3^{\alpha}\cdot B^{\gamma} \tag{2} $$
Note that $\alpha$ is odd and $B$ is not necessarily a prime factor.
We've already argued $x\not\equiv 0\pmod{3}$, so $x\equiv +1\pmod{3}$ or $x\equiv -1\pmod{3}$.
If $x\equiv +1\pmod{3}$ then $x^2-1=3^{2\lambda}\cdot D^\eta$  and $x^2+1\not\equiv 0\pmod{3}$. So $\alpha\ne2\lambda$.
If $x\equiv -1 \pmod{3}$, then $x^2+1=3^{2\lambda}\cdot D^\eta$  and
$x^2-1\not\equiv 0\pmod{3}$. So $\alpha\ne2\lambda$.
Hope this helps.
