Of course, $x=0$ is an answer, so let's look for non-zero ones from now on. If the given expression is a perfect square, so is
four times the number, which is $4x^4+4x^3+4x^2+4x+4$.
Now we try to estimate it by "nearby" perfect squares.
In particular, the square of $(2x^2+x)$ is $4x^4+4x^3+x^2$, which is "obviously" too small. But the square of $(2x^2+x+2)^2$ is $4x^4+4x^3+9x^2+4x+4$, and that's "obviously" too big. Thus the only possibility is the perfect square between them, $(2x^2+x+1)^2$. Solving the equation $(2x^2+x+1)^2=4x^4+4x^3+4x^2+4x+4$ by usual methods, we get $x=3$ and $x=-1$ as the only roots.
One could rightly object against the "obviously"s above since we are dealing with possibly negative numbers here. Fortunately, this is easily settled:
We see that $$(4x^4+4x^3+4x^2+4x+4)-(2x^2+x)^2=\frac13 (3x+2)^2+\frac83>0,$$ and that $$(4x^4+4x^3+4x^2+4x+4)-(2x^2+x+2)^2=-5x^2<0$$for non-zero $x$, so the above bounds are all legal.
Thus all the possible $x$'s are
0,-1, and 3.