A number theory problem by Prithwish Roy

$2x^4+1402=y^4 \\ => 2(x^4+701)=y^4$

L.H.S. contains only a factors of $2$ so $2 | y^4$ .So, $y=2y_1$

Substituting this in the equation $2(x^4+701)=16 y_1^{4} \implies x^4+701=8y_1^4$ . Right side is even then left side should be also even.Then $x$ should be always odd.Let's assume that $x=2x_1+1$ .Putting this in the equation

$(2x_1+1)^4 + 701=8y_1^4$

Expanding in the form of bini\omial

$16 x_1^4 + \binom{4}{1}.8.x_1^3+\binom{4}{2}.4.x_1^2+\binom{4}{3}2.x_1+1+701=8y_1^4 \\ 8k+702=8y_1^4$

As $702$ is not divisible by $8$ that's why there is no integal solution.