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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 $b$.