Solving a system of Pell and Pell-like equations Solve (Find all the solutions, if there are, or prove there are not)  the system of 2 Pell or Pell-like simultanous equations over the positive integers :
$2b^{2}= a^{2} +1 = 3k^{2} + 2 $ with the 3 variables (a,b,k).
Note that the 2 equations are :
 $a^{2} -2b^{2} = -1$ and  $a^{2} -3k^{2} = 1$
The third one: $ 2b^{2} -3k^{2} =2$ does not need to be solved if the other two are.
 A: László Szalay, On the resolution of simultaneous Pell equations, Annales Mathematicae et Informaticae 34 (2007) pp. 77–87, is available here. I don't know whether the precise pair of equations you give is discussed in the paper, but it does give you the tools and references you need to approach this kind of problem. 
EDIT: Having looked a bit more closely at the paper, I believe a solution to your problem is in fact given in the paper and attributed to Riele, reference 24. 
A: Here is the solution to the simultanous
$$
\begin{align}3x^2−y^2=2\tag{1}\\8x^2−z^2=7\tag{2}\end{align}
$$
$$
x_{n+1}=2x_n+y_n\quad\text{and}\quad y_{n+1}=3x_n+2y_n
$$
for $(1)$ and
$$
x_{n+1}=3x_n+z_n\quad\text{and}\quad z_{n+1}=8x_n+3z_n
$$
for $(2)$. And we get: $(1)\Rightarrow(x,y)=(1,1);(3,5);(11,19);(41,71);(153,265);(571,989)$ and $(2)\Rightarrow(x,z)=(−1,1);(2,5);(11,31);(64,181);(373,1055);(2174,6149)$. The solutions coincide in the same $n^{\text{th}}$ position, the third position: for the only solution $(x,y,z) = (11,19,31)$.
Clearly, x grows faster, after its values coincide, permanently in (2) than in (1), and thus we prove this is the only non trivial solution. The case (2) has another family of solutions : $(1,1);(4,11);(23,65);(134,379)$ which obviously still grow faster, beginning with a higher value of one of its initial terms and will never give a solution.
A: $a^2 - 3*k^2 = 1$ (1) , general solution: 
$a(n+1) = 2a(n) + 3k(n)$, $k(n+1) = a(n) + 2k(n)$. The first 6 solutions, without counting the  trivial solution: (1,0) are:
$(a,k) = (2,1); (7,4); (26,15); (97,56); (362,209); (1351,780);\dots$
$a^2 - 2*b^2 = -1$ (2), general solution: 
$a(n+1) = 3a(n) + 4b(n)$, $b(n+1) = 2a(n) + 3b(n)$. The first 6 solutions are:
$(a,b) = (1,1); (7,5); (41,29); (239,169); (1393,985); (8119,5741);\dots$
Once we match the same a value in both equations: a=7 asociated to k=4 and b=5; we do not need to look any  more for solutions, since both a and b grow faster in one equation than a and k in the other ---> the only solution non trivial is (a,b,k) = (7,5,4). The problem would be if a(n) grew faster and b(n) slower in one equation than in the other, or the contrary. In that case we would have to match another coincident value of "a" among the infinite values in both equations. But no problem here.
The general solution of the third equation $2b^{2} -3k^{2} = 2$ is $b(n+1) = 5b(n)+ 6k(n) ; k(n+1) = 4b(n) + 5k(n)$ with solutions: $(1,0);(5,4);(49,40);(485,396);(4801,3920), ...$
