Solve for system of diophantine equations $\cases{x+1=a^2 \cr x^3-x^2+1=b^2}$

I just can found a trivial solution $x=0$. Is there any other ?

Here's an elementary solution.  First note that we may assume $a,b \ge 0$ since negative values yield exactly the same $x$.  It's easy to see that $a=0$ doesn't work and $a=1$ does, so let's assume $a \ge 2$.
The two equations are obviously equivalent to $(a^2-1)^3 - (a^2-1)^2 + 1 = b^2$.  The LHS expands to $a^6 - 4a^4 + 5a^2 - 1 = (a^3-2a)^2 + (a^2-1)$.
Since $a \ge 2$, both $a^3-2a$ and $a^2-1$ are positive.  In general, whenever $n^2+m$ is a square ($n,m$ positive integers) we must have $m \ge 2n+1$ by considering the smallest square after $n^2$.
Therefore, if $(a^3-2a)^2 + (a^2-1)$ is a square, $a^2-1 \ge 2(a^3-2a)$.  But it's easy to check this is impossible for $a \ge 2$, so there are no further solutions.
This simple technique is often very effective for detecting square values of monic even-degree polynomials in general.
