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 $(11,23).$

Try working mod $3$ and mod $8$. Assuming $x, y>3$, we have $x,y = \pm 1$ mod $3$. Since $x, y$ are odd we have $x^2, y^2=1$ mod $8$, so
$$x^2, y^2 = 1 \text{ mod } 24.$$
Substituting in the equation gives $$x = 24k+11 $$ for some integer $k$.
Rearranging the original equation we get
$$x(5x-7)=(y-1)(y+1), \tag{1}$$
therefore $x |y-1$ or $x|y+1$, since $x$ is a prime number.
Solving for $x$ gives
 $$ x = \frac{17}{10} + \frac{1}{10}\sqrt{20y^2+29}>\frac{1}{3}(y+1).$$
Note that $x$ is odd and $y \pm 1$ is even, so $x \ne y\pm1$. This forces $x = \frac{1}{2} (y \pm1)$, or 
$$y = 2x \pm 1 = 48k + 22 \pm 1 \Rightarrow y = 48k+23.$$
$48k+21$ is rejected being divisible by 3.
Plugging these in $(1)$ gives the solution $k=0$ or
$$x = 11, \space y = 23.$$
