System of Diophantine Equations I'm working on this problem I came across on the internet but I have no solution yet. 
The problem states: Find all prime numbers p that are such that  $p+1=2x^2$ and $p^2+1=2y^2$ where x and y are integers.
Edit: to which point I have reached.
 I wrote $p=2p_1+1$. After this the second equation becomes $2p_1^2+2p_1+1=y^2$ which is rewritten as $p_1^2+x^4=y^2$ in which we only have to find Pythagorean triples. 

$$p+1=2x^2,p^2+1=2y^2\\
2y^2-1=(2x^2-1)^2\\
y^2=2x^4-2x^2+1\\
y^2=2x^2 (x-1)(x+1)+1$$
Suppose $y=2m+1,$ then $$y^2=4m(m+1)+1.$$
Hence 
$$2m(m+1)=x^2 (x-1)(x+1).$$
$$\dfrac{m(m+1)}{2}=\dfrac{x(x-1)}{2}\times\dfrac{x(x+1)}{2}$$
I have no idea to continue this. But I think find the pairs triangle numbers who can satisfy the equation $$\color{Green}{T_m=T_{x-1}T_x}$$ would be a help to solve tyhis problem.
