<s>[INST] Perfect Square relationship with no solutions I would like to show that for positive integers $a>b,c$ all greater than 1 such that $c\nmid a$, there are no solutions to the following equation:
$$a^2+1=b^2(c^2+1)$$
As was pointed out in the comments, it might help to rearrange and factor:
$$(a-b)(a+b)=(bc+1)(bc-1)$$
Another interesting approach to the question, if we rewrite:
$$a^2-b^2c^2=b^2-1$$
Then we are trying to argue that $b^2-1$ cannot be squarefree (in particular, $c^2|b^2-1$)
 [/INST] No. On the other hand, it appears that $c | a.$
I am watching a longer computer run, one infinite family turns out to be 
$$   a = 4 c^3 + 3 c, \; \; \;  b = 4 c^2 + 1$$
for any positive integer $c.$ You can check that $a^2 + 1 = b^2 (c^2 + 1).$
There are a few others, though, with especially small $c,$ which makes me think they might be derivable from the infinite sequence. 
EEDDIITTTTT: that worked. I have a doubly infinite sequence, one infinite sequence for a given $c.$ 
$$   a_0 =  c, \; \; \;  b_0 =  1,   $$
$$   a_1 = 4 c^3 + 3 c, \; \; \;  b_1 = 4 c^2 + 1,$$
$$   a_2 = 16 c^5 + 20 c^3 + 5 c , \; \; \;  b_2 = 16 c^4 + 12 c^2 + 1,$$
then
$$ \color{magenta}{  a_{n+2} = (4 c^2 + 2) a_{n+1} - a_n, \; \;  b_{n+2} = (4 c^2 + 2) b_{n+1} - b_n}.  $$
I believe  already that this doubly indexed family gives all solutions.

 a     b    c
38    17    2     
117    37    3     
268    65    4     
515    101    5     
682    305    2     
882    145    6     
1393    197    7     
2072    257    8     
2943    325    9     
4030    401    10     
4443    1405    3     
5357    485    11     
6948    577    12     
8827    677    13     
11018    785    14     
  a       b      c


</s>