Let us assume there are integers $a$, $b$, $M$ and $N$ such that $a + b = M^2$ and $ab - 1 = N^2$.

First case: The numbers $a$ and $b$ are even.
From $a \cdot b = N^2 + 1$ we conclude that $N^2 + 1$ is divisible by $4$ which is impossible.

Second case: Exactly one of the numbers $a$ and $b$ (say $a$) is even.
From $a \cdot b = N^2 + 1$ we conclude $b = 4k + 1$ for some integer $k$, because $N^2 + 1$ has no divisors of the form $4k + 3$ for an integer $k$. Also, $N$ is odd so we can write it in the form $N = 2l + 1$ for some integer $l$. From $ab = (2l + 1)^2 + 1 = 4l^2 + 4l + 2$ we conclude that $a = 4m + 2$ for some integer $m$. Then $a + b = 4n + 3$, for some integer $n$, so it can not be a perfect square. We got a contradiction.

Third case: Both numbers $a$ and $b$ are odd.
Then the number $M$ is even, so $M^2$ is divisible by $4$. Hence one of the numbers $a$ and $b$ has to have residue $3$ when divided by $4$. We again got a contradiction since $N^2 + 1$ can not have a divisor of the form $4k + 3$ for some integer $k$.

Since all cases lead to a contradiction we conclude there are no such $a$ and $b$.