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

This curve $y^2=3x^4-3$ has only two integral points, according to the magma online calculator as follows.
IntegralQuarticPoints($[3,0,0,0,-3]$);
It says that all integral points are $( \pm 1 , 0)$.
Hence this equation has no integer $(y\neq 0)$ solution.
See related info MO 
