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

Solve $$7^x+x^4+47=y^2$$ where $x, y \in \mathbb{N}^*$

If $x$ is odd then the left term is congruent with $3$ mod $4$ so it couldn't be a perfect square, so we deduce that $x=2a$ and the relation becomes $$49^a+16a^4+47=y^2$$ and it is easy to see that the left term is divisible by $16$ so we obtain that $y=4b$, so we have to find $a$ and $b$ such that $$49^a+16a^4+47=16b^2$$From this point I was completely stuck. I think that there are no solutions but how can I prove it?

Solution 1:

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 $6$.

Solution 2:

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