, 2015

If $a=1$ or $b=1$ we obtain the solutions $(1, 2^n-1)$ and $(2^n-1, 1)$, with $n>1$.

Let $a,b \ge 2$ and $a<b$. Note that then $a+b < ab+1$ due to the identity $(ab+1)-(a+b) = (a-1)(b-1)$. Let $a+b = 2^n$, $n \ge 2$; in fact then $n \ge 3$ as $n=2$ forces $a=b=2$. Then $a = 2^{n-1}-c$, $b = 2^{n-1}+c$ with $1 \le c < 2^{n-1}$. We have
$$2^n = a+b < ab+1 = 2^{2(n-1)}-c^2+1 \le 2^{2(n-1)},$$
which implies $ab+1 = 2^k$ with $n+1 \le k \le 2(n-1)$. Write
$$2^k = 2^{2(n-1)} - c^2 + 1,$$ i.e., $ab+1 = 2^k$, in the form
$$(c-1)(c+1) = 2^{2(n-1)} - 2^k.$$ Then $(c-1)(c+1)$ is divisible by $2^k$ since $k \le 2(n-1)$. On the other hand $n+1 \le k$, so $(c-1)(c+1)$ is divisible by $2^{n+1}$. Clearly $c-1$ and $c+1$ are consecutive even numbers. Since one of them is divisible by $2$ but not by $4$, the other one is divisible by $2^n$.
Now the inequalities $1 \le c < 2^{n-1}$ show that the later is possible only if $c-1=0$, i.e., $c=1$. Hence $a = 2^{n-1}-1$, $b = 2^{n-1}+1$ with $n \ge 3$. This is clearly an admissible pair.
Hence the solutions are $(1, 2^n-1)$, $(2^n-1, 1)$ and $(2^n-1, 2^n+1)$, $(2^n+1, 2^n-1)$ with $n>1$.