Finding specific values of $n$ such that $n^2+3n+3$ is factorisable with certain constraints on the factors

Let $c:=b-a$, so that $|c|<2\sqrt{n+1}$ and $a$ and $-b$ are roots of $$X^2+(b-a)X-ab=X^2+cX-(n^2+n+3).$$ Because this quadratic has two integral roots its discriminant $\Delta$ is a perfect square, where $$\Delta=c^2+4(n^2+3n+3).$$ Of course $c^2\geq0$ so $$\Delta\geq4(n^2+3n+3)=(2n+3)^2+3,$$ which shows that $\sqrt{\Delta}>2n+3$. On the other hand, because $c<2\sqrt{n+1}$ it follows that $$\Delta<4(n+1)+4(n^2+3n+3)=4(n^2+4n+4)=(2n+4)^2,$$ which shows that $\sqrt{\Delta}<2n+4$, a contradiction. Hence it is never possible.

As you suggested, WLOG, let $a \le b$, with an integer $c$ such that

$$b = a + c \tag{1}\label{eq1A}$$

The constraint then means

$$0 \le c \lt 2\sqrt{n + 1} \implies 0 \le c^2 \lt 4(n + 1) \tag{2}\label{eq2A}$$

We then get

$$\begin{equation}\begin{aligned} a(a + c) & = n^2 + 3n + 3 \\ a^2 + ac & = n^2 + 3n + 3 \\ 4a^2 + 4ac & = 4n^2 + 12n + 12 \\ 4a^2 + 4ac + c^2 & = 4n^2 + 12n + 12 + c^2 \\ (2a + c)^2 & = 4n^2 + 12n + 12 + c^2 \end{aligned}\end{equation}\tag{3}\label{eq3A}$$

Using \eqref{eq2A} gives

$$4n^2 + 12n + 12 \le (2a + c)^2 \lt 4n^2 + 12n + 12 + 4(n + 1) = 4n^2 + 16n + 16 \tag{4}\label{eq4A}$$

Note $(2n + 3)^2 = 4n^2 + 12n + 9$ is less than the lower boundary. Also, $(2n + 4)^2 = 4n^2 + 16n + 16$ is the exclusive upper boundary. This means there's no perfect square in that range, i.e., there's no integer $2a + c$ which satisfies \eqref{eq4A} and, thus, no integer $n$ matching the conditions.