We show that, if some integer $n \ge 2$ satisfies the required condition, then so does $n^2 + a - 1$. Since $n^2 + a - 1 > n$, it is then sufficient to exhibit a single $n \ge 2$ that fits the bill.
Consider the identity
$$(n^2 + a - 1)^2 + (n^2 + a - 1) + a = (n^2 + n + a)(n^2 - n + a). \quad (*)$$
Suppose now that $n!$ is divisible by $n^2 + n + a$ for some integer $n \ge 2$. Since $n < n^2 - n + a < n^2 + a - 1$, it follows that $(n^2 + a - 1)!$ is divisible by $n!(n^2 - n + a)$; and since $n!$ is divisible by $n^2 + n + a$, the identity (*) then shows $(n^2 + a - 1)!$ divisible by $(n^2 + a - 1)^2 + (n^2 + a - 1) + a$.
Next, set $n = a$ in (*) to get $(a^2 + a - 1)^2 + (a^2 + a - 1) + a = a^3(a + 2)$. If $a \ge 3$, then $a < a + 2 < a^2 < a^2 + a - 1$, so $a^2 + a - 1 \ge 2$ and $(a^2 + a - 1)!$ is divisible by $a^3(a + 2) = (a^2 + a - 1)^2 + (a^2 + a - 1) + a$.
If $a = 2$, set $n = 10$, and notice that $n! = 10!$ is divisible by $2 \cdot 7 \cdot 8 = 112 = 10^2 + 10 + 2 = n^2 + n + 2 = n^2 + n + a$.
Finally, if $a = 1$, set $n = 16$, and notice that $n! = 16!$ is divisible by $3 \cdot 7 \cdot 13 = 273 = 16^2 + 16 + 1 = n^2 + n + 1$.

Alternative solution:
The argument hinges on the fact that the equation $u^2 - 4a(a+2)v^2 = 1 - 4a$ has infinitely many solutions $(u_k, v_k)$ in positive integers, where the $u_k$ and the $v_k$ both form strictly increasing sequences.
Assume this for the moment. Clearly, the $u_k$ are all odd, so the $n_k = \frac{1}{2}(u_k - 1)$ are all integers. Consider first the case $a \ge 2$. Since the $u_k$ form a strictly increasing sequence of positive integers, so do the $n_k$, and, since $a \ge 2$,
$$n_k = \frac{1}{2}(u_k - 1) > \left( \frac{u_k^2 + 4a - 1}{a(a + 2)} \right)^{\frac{1}{2}} = 2v_k,$$
for all but finitely many indices $k$.
Recalling that the $v_k$ also form a strictly increasing sequence of positive integers, it follows that the numbers $a(a+2) < v_k < 2v_k$ all three occur in $n_k!$, so the latter is divisible by $a(a+2) \cdot v_k \cdot 2v_k = 2a(a+2)v_k^2 = \frac{1}{2}(u_k^2 + 4a - 1) = \frac{1}{2}(u_k - 1)^2 + (u_k - 1) + 2a = 2(n_k^2 + n_k + a)$, for all but finitely many indices $k$, as desired.

Finally, consider the equation $u^2 - 4a(a+2)v^2 = 1 - 4a$ along with the associated Pell equation $x^2 - 4a(a+2)y^2 = 1$. The pair $(u_0, v_0) = (2a+1, 1)$ solves the former and the pair $(x_0, y_0) = (2(a+1)^2 - 1, a+1)$ solves the latter. For each positive integer $k$, let $u_k = u_0x_k + 4a(a+2)v_0y_k$ and let $v_k = v_0x_k + u_0y_k$, where $x_k = x_0x_{k-1} + 4a(a+2)y_{k-1}$ and $y_k = y_0x_{k-1} + x_0y_{k-1}$. Since the $(x_k, y_k)$ solve the Pell equation, the $(u_k, v_k)$ solve the equation under consideration. Clearly, the $x_k, y_k, u_k, v_k$ all form strictly increasing sequences of positive integers. This completes the proof for $a \ge 2$.
The case $a=1$ is dealt with similarly by considering the equation $u^2 - 28v^2 = -3$ along with the associated Pell equation $x^2 - 28y^2 = 1$. The pair $(u_0, v_0) = (5, 1)$ solves the former and the pair $(x_0, y_0) = (127, 24)$ solves the latter. The rest of the argument goes through verbatim.