Thanks by your help I found a solution and I will leave it here if someone finds this problem interesting
$$\begin{cases}ab+1&=&(c+1)(d+1)\\ cd+1&=&(a-1)(b-1)\end{cases}$$
When you expand the both sides of the equations you get: $$\begin{cases}ab+1&=&cd+c+d+1\\ cd+1&=&ab-a-b+1\end{cases}$$
Sum both equations and you get: $$\begin{cases}a+b&=&c+d\\ ab-cd&=&a+b\end{cases}$$
From here it’s obvious that
ab-cd=a+b=c+d and let’s consider these sums equal to s
Then
s=a(s-a)-c(s-c)=(a-c)(s-a-c)
If s is odd then a-c & s-a-c are also odd but then their sum s-2c would be even and so s would be even too
Therefore a-c and s-a-c are even & let’s consider a-c=2x and s-a-c=2y where x,y are integers
From this we get that s=4xy and then we’re done
We get (a,b,c,d)=(2xy+x-y,2xy-x+y,2xy-x-y,2xy+x+y)
