Prove that $x^n=2y^2+3$, $n\equiv3\pmod{4}$, has no solution $(x,y,n)$ in positive integers.
I managed to get $x\equiv3\pmod{8}$ and $y$ is even and $n\equiv7\pmod{8}$. After taking modulo $2$, $x$ has to be odd.
Note that $x\equiv x^n\equiv 2y^2+3\equiv 3,5\pmod{8}.$
Also $x^n-1=2(y^2+1)$, and the odd prime factors of $y^2+1$ are $1\pmod{4}$. If $x\equiv 1\pmod{4}$, then $n\equiv x^{n-1}+...+1\equiv 1\pmod{4}$. $\bot$
So $x\equiv 3\pmod{8}$. Then $2y^2+3\equiv x^n\equiv 3\pmod{8}$, so $y$ is even.
COMMENT.-It is not hard to prove that the odd $x\equiv1,5,7\pmod{10}$ and $x\not\equiv3,7,9\pmod{10}$ so if you follow this reduction modulo $10$ you need to prove the impossibility with $n\equiv3\pmod4$ of $$(10x+1)^n=2y^2+3\Rightarrow y\equiv2,3,7,8\pmod{10}\hspace 1cm (1) \\(10x+5)^n=2y^2+3\Rightarrow y\equiv 1,4,6,9\space\space\pmod{10}\hspace 1cm (2)\\(10x+7)^n=2y^2+3\Rightarrow y\equiv5\pmod{10}\hspace 1cm (3)$$ This is a set of nine possible equations to prove be impossible. You must take into account that for exponent $n$ large and not prime, diophantine equations are hard to solve.