For $x^2-3y^2=1$ over integers more than $1$, can $\frac{y+1}2$ be square number? 
For $x^2-3y^2=1$ over integers more than 1, can $\frac{y+1}2$ be square number?

I know that $x^2-3y^2=1$ is one of pell's equation, so I know its general solution. But I know nothing about its properties, and I can't proceed my proof. How should I approach this question?
 A: $$x^2-3y^2=1\implies y^2=\frac{x^2-1}{3}\implies \frac{x-1}{p}\cdot\frac{x+1}{q}\quad\text{where}\quad p,q\quad \text{ divides }\quad 3$$
It is easy to see the solutions of $p=1,q=3\text{ and }x=1\lor x=2$ but perhaps there are other values of x divisible by these factors and, it happens there are.
For $x-1$, x can be any integers and, $x+1$ can be any multiple of $3$ such as $3,6,9$ but the result, divided by 3 must be a perfect square and these get rarer with altitude. Here is a sample of infinite $(x,y)$ solutions. Only positive integers are shown for simplicity but negatives apply as well.
$$(x,y)\in\{(1,0),(2,1),(7,4),(26,15),(97,56),(362,209), \cdots\}$$
This gives no definition of the set. A search is still required and although a solution for $x$ would be faster it alone yields just few insights into the values of $y$ that yield integers.
$$x^2-3y^2=1\implies x^2={3y^2+1}$$
There is a brighter side though, in that both $x$ and $y$ values are known sequences in the Online Encyclopedia of Integer Sequences.
Sequence A001075
shows $x\in\{ 1, 2, 7, 26, 97, 362, \cdots\}$
Sequence A001353
shows $y\in\{ 0, 1, 4, 15, 56, 209, \cdots\}$
These sequences often come with several formulas for their generation and perhaps one of them may meet your needs in generating the $n^{th}$ pair directly.
