Find all prime solutions of equation $5x^2-7x+1=y^2.$ Find all prime solutions of the equation $5x^2-7x+1=y^2.$
It is easy to see that 
$y^2+2x^2=1 \mod 7.$ Since $\mod 7$-residues are $1,2,4$ it follows that $y^2=4 \mod 7$, $x^2=2 \mod 7$  or $y=2,5 \mod 7$ and $x=3,4 \mod 7.$ 
In the same way from $y^2+2x=1 \mod 5$ we have that $y^2=1 \mod 5$ and $x=0 \mod 5$ or $y^2=-1 \mod 5$ and $x=4 \mod 5.$
How put together the two cases?
Computer find two prime solutions $(3,5)$ and $(11,23).$

Since $(0,1)$ is a solution to $5x^2-7x+1=y^2$, it  can be  used  to parametrize all rational solutions to $5x^2-7x+1=y^2$. That will give us:
$$x:=\frac{-2ab - 7b^2}{a^2 - 5b^2}$$
and
$$y:=\frac{a}{b}x+1$$
where $a,b\in \mathbb{Z}$, $b\neq 0$.
Since $x$ is prime, it follows that either $b=1$ or $x=b$.

*

*case $b=1$
This gives us $x:=(-2a - 7)/(a^2 - 5)$. Since $x\in \mathbb{Z}$,
we get  $a=\pm 2$, and then $x=11$ or $x=3$.
Those will give $y=23$ or $y=5$, repectively.


*case $x=b$.
We have  that $\frac{-2ab - 7b^2}{a^2 - 5b^2}=b$ gives
$$(*) \hspace{2cm} a(a + 2) =5b^2 - 7b.$$
Since $y=a+1$ and $x=b$ are prime, and  $x=2$ or $y=2$ do not give solutions to $5x^2-7x+1=y^2$, we conclude  that  $y=a+1$ and $x=b$ are ODD primes. In particular, $a(a + 2)\equiv 0 \mod 4$.
Now reducing (*) $\mod 4$, contradicts the fact that $b$ is odd. Thus, the case $x=b$ does not occur.
Therefore  $(x,y)= (11,23)$ and  $(x,y)=(3,5)$ are the only solutions
where both coordinates are prime numbers.
