Solve $7^x+x^4+47=y^2$

If $x$ is even and let $x=2a$, then $(7^a)^2<7^x+x^4+47<(7^a+1)^2=7^x+2\times7^a+1$ if $(2a)^4+47<2\times7^a+1$, which is true for $a \ge 4$. Therefore, it is enough to consider only $x=2, 4$ and $6$.

You can't prove no solutions. $x=4, y=52$ is a solution.