Pretty ugly solution during contest: Let us take the equation $(mod p)$ to obtain $p |x^2-y^2$ Now we consider the cases for $p |x-y$ or $p |x+y$ Case 1: $p |x-y$ We may rewrite the equation as $p^2+2p=(y(p+1)-x)(y(p+1)+x)$ Note that $y(p+1)-x \equiv y-x \equiv 0 (mod p)$ Thus $y(p+1)-x \ge p$ since both $p^2+2p , y(p+1)+x$ are positive, which gives: $$p+2 \ge y(p+1)+x$$which is a contradiction since $x,y \ge 2$ Case 2: $p |x+y$ Also we have $y(p+1)+x \equiv y+x \equiv 0 (modp)$. Let $y(p+1)+x=kp$ for some $k \in \mathbb{N}$ Then since $p^2+2p=(y(p+1)-x)(y(p+1)+x)$ $\implies$ $y(p+1)-x=\frac{p+2}{k}$ Solving the system of equation for $x,y(p+1)$ we get $x,y$ as: $$x=\frac{k^2p-(p+2)}{2k}, y=\frac{k^2p+(p+2)}{2k(p+1)}$$Note that this implies that $k |p+2$ and $2k(p+1) |k^2p+(p+2)$ Then since $2k(p+1) |2k^2(p+1)=2k^2p+2k^2$, Thus; $$2k(p+1) |(2k^2p+2k^2)-2(k^2p+(p+2))=2k^2-2p-4$$Now we consider the three subcases: Case 1: $2k^2-2p-4=0$ We have $k^2=p+2$ then: $y=\frac{k^2p+(p+2)}{2k(p+1)}=\frac{p+2}{2k}$ $\implies$ $p=2$, which doesn't give a solution for x. case 2: $2k^2-2p-4<0$ We have: $$2(p+1) \le 2k(p+1) \le 2p+4-2k^2 \le 2p+2$$$\implies$ $k=1$ which gives $x<0$ case3: $2k^2-2p-4>0$ $k |p+2$ gives $k-2\le p$ $$ 2k(k-1)\le 2k(p+1) \le 2k^2-2p-4 $$$\implies$ $p+2 \le k$ since the reverse inequality holds we have $k=p+2$, Thus plugging back for $x,y$: $(x,y)=(\frac{p^2+2p-1}{2},\frac{p+1}{2})$ which also works.

The solution is $(x,y)=\boxed{(\frac{p^2+2p-1}{2},\frac{p+1}{2})}$. First notice that $p=2$ yields no solutions. Multiplying and factoring we get, $$p(p+2)=(yp+y)^2-x^2$$Now let $a$ and $b$ be natural numbers such that $ab=p+2$ and that either $y(p+1)-x=ap$ and $y(p+1)+x=b$ or $y(p+1)-x=a$ and $y(p+1)+x=bp$. Adding the equations together we get $2y(p+1)=ap+b$ or $2y(p+1)=a+bp$ taking modulo $p+1$ we get that $a-b\equiv 0$. By size constraints since $ab=p+2$ we must have $(a,b)=(1,p+2),(p+2,1)$ which can easily be checked to only produce the desired solution or we could have $a=b=\sqrt{p+2}$. Then $2y(p+1)=\sqrt{p+2}(p+1)$ this implies that $\sqrt{p+2}$ is even, a contradiction.

https://drive.google.com/file/d/10LgS9xCTwHBmYG-ubTJCbfNGxz1qGk1s/view?usp=drivesdk it's my solution.

actually i dont know vieta jumping now but its more easy with vieta jumping one of my friend solved it with vieta jumping

How can we use vieta jumping? Please tell me.