$4y^2=16x^2+808x=(4x+101)^2-101^2$ $101^2=(4x+101-2y)(4x+101+2y)$ As $101$ is prime than $4x+101+2y=101^2,4x+101-2y=1 \to x =1250,y=2550$

We have $4y^2 = (4x+101)^2 - 101^2$. Therefore, $$(4x+101 -2y)(4x+101 +2y) = 101^2.$$Hence, $4x + 101 - 2y = a, \ 4x+ 101 +2y = b$, where $ab = 101^2$. Also, $\displaystyle x = \frac 14 \left(\frac{a+b}{2} - 101 \right)$ and $\displaystyle y = \frac{b-a}{4}$, therefore $b > a$ since $x,y$ are positive integers. Now, $a<b$ implies that $a < 101$. Thus, the only pair $(a,b)$ is $(1,101^2)$ since $101$ is prime. Thus $x = 1250$ and $y = 2550$. $\blacksquare$

Multiplying by $4$ and completing the square, we get, $(4x + 101)^2 - 4y^2 = 101^2$. Which means $(4x + 101 + 2y, 4x + 101 -2y) = (101^2, 1)$. This gives $(x, y) = (1250, 2550)$. $\blacksquare$