Solve $x^2+3y = u^2$ and $y^2+3x=v^2$ in positive integers.

The question is from the pg – 59 from ‘ An Introduction to Diophantine Equations ‘ by Titu Andreescu , Dorin Andrica , Ion Cucurezeanu.

Example 1 : Solve in positive system of equations in positive integers

$$begin{cases} x^2+3y = u^2 \ y^2 + 3x = v^2 end{cases}$$
$;;;;;;;;;,,,,,,,,,,,,text{(Titu Andreescu)}$

Solution. The inequality $x^2 + 3y ≥ (x + 2)^2 , y^2 + 3x ≥ (y + 2)^2$
cannot both be true, because adding them would yield a contradiction.

So at least one of the inequalities $x^2 + 3y < (x + 2)^2$ and
$y^2 + 3x < (y + 2)^2$ is true. Without loss of generality, assume
that $x^2 + 3y < (x + 2)^2$.

Then $$x^2 < x^2 + 3y < (x + 2)^2 implies
x^2 + 3y = (x+1)^2$$ or, $3y = 2x+ 1$ . We obtain $x = 3k + 1, y = 2k + 1$
for some nonnegative integer $k$ and $y^2 + 3x = 4k^2 + 13k + 4$.

For
$k > 5, (2k+ 3)^2 < 4k^2 + 13k+ 4 < (2k+ 4)^2$ ; hence $y^2 + 3x$ cannot be
a perfect square. Thus we need only consider $k ∈ {{0, 1, 2, 3, 4}}$ . Only
$k = 0$ makes $y^2 + 3x$ a perfect square; hence the unique solution is
$$x = y = 1,;;;;;; u = v = 2.$$

But if we take , $$4k^2+13k + 4 = v^2$$ $$implies k = dfrac{-13 pmsqrt{105+16v^2}}{8}$$

Since $105+16v^2 = a^2 implies 105 = (a-4v)(a+4v)$ which gives $a in {pm11 , pm13 , pm19 ,pm53}$ .
Out of these , only $a in { pm13 , pm53}$ works which gives $k=0,5$ , And so the the answer should be $$(x,y,u,v) = (1,1,2,2);;;,(16,11,17,13);;;;,(11,16,13,17)$$

Who is correct here?