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?

When $x$ is odd then $7^x+x^4+47\equiv 3(\mod 4)$. So, it is not possible $y^2\equiv 3(\mod 4)$.
Let $x=2k$, then for $k\geq 4$
$$(7^k)^2<7^{2k}+(2k)^4+47<(7^k+1)^2$$
This mean we have $k\leq 3$.
if try to $k=1,2,3$ only $x=4$ is a solution.
