How are such sets of natural numbers called? I heard about this problem an year ago, but I just can't remember the name.
The problem goes like this: study the sets 

$\{a_1,a_2,\dotsc,a_m\}\subseteq\mathbb{N}$ such that if $1\leq i<j\leq m$, then $a_i a_j+1$ is a perfect square.

Is there a technical term for such sets?
 A: Sets of $m$ integers with this property are called integer Diophantine $m$-tuples. A good starting point to learn about them (and about rational Diophantine $m$-tuples) is this paper by 
Dujella, Kazalicki, Mikic, and Szikszai.
A: I will tell you about some special cases.
$$\left\{\begin{aligned}&ab+T=x^2\\&ac+T=y^2\\&bc+T=z^2\end{aligned}\right.$$
We need to find solutions $a,b,c$ - that was an arithmetic progression.  This will help the solution of the equation Pell.
$$p^2-3s^2=T$$
Knowing any solution of the equation Pell $(p_0;s_0)$ you can always find the following formula.
$$p_1=2p_0+3s_0$$
$$s_1=p_0+2s_0$$
Having any decision - can immediately write down the solution of this system.
$$a=2s-p$$
$$b=2s$$
$$c=2s+p$$
$$x=s-p$$
$$y=s$$
$$z=s+p$$
Interesting here is that the $x,y,z$ looks like an arithmetic progression.
I thought as this task to generalize and use for any numbers.  It turned out that you can do without calculations.  For the system of equations:
$$\left\{\begin{aligned}&ab+T=x^2\\&ac+T=y^2\\&bc+T=z^2\end{aligned}\right.$$
Enough to factor the following number:  
$$bc=(y+c)^2-T$$
Using these numbers you can easily write the solution of this system of equations.
$$a=b-c-2y$$
$$b=b$$
$$c=c$$
$$x=b-c-y$$
$$y=y$$
$$z=y+c$$
It is better to use a more General approach. We write the system.
$$\left\{\begin{aligned}&xy+T=a^2\\&xz+T=b^2\\&xq+T=c^2\\&yz+T=d^2\\&yq+T=k^2\\&zq+T=n^2\end{aligned}\right.$$
If the number $T$, lay at the multipliers. We find then the desired settings.
$$T=3(p-t-s)(p+t-s)(p+s)^2$$
Then the solution can be written as.
$$x=t^2+2s^2+2ps-p^2$$
$$y=t^2-s^2+2ps+2p^2$$
$$z=4t^2-(p-s)^2$$
$$q=3(p+s)^2$$
$$a=p^2+ps+s^2-t^2$$
$$b=2t^2+s^2+ps-2p^2$$
$$c=3(p+s)s$$
$$d=2t^2-2s^2+ps+p^2$$
$$k=3(p+s)p$$
$$n=3(p+s)t$$
