We first note that, independently of the value of $A$, $n = 0$ implies $k = 0$ and vice versa. Furthermore, also independently of the value of $A > 0$, there can be no solution for either $n = 1$ or $k = 1$, since $n! + A \cdot n > n^k$ certainly holds in either of these cases. In the following, we therefore limit our discussion to the case $n, k \ge 2$. In this case we can divide the given equation by $n$, which yields the equivalent equation
$$(n - 1)! + A = n^{k-1}.$$
We now turn our attention to the case $A = 7$.
If $n = 2$, the equation yields $1 + 7 = 2^{k-1}$, which is true for $k = 4$, yielding the solution $(n, k) = (2, 4)$. If $n > 2$, $(n - 1)!$ is even, and $n^{k-1}$ and therefore $n$ must be odd. If $n = 3$, we have $(3 - 1)! + 7 = 3^{3-1}$, and we see that $(n, k) = (3, 3)$ is another solution. We now wish to show that there are no others.
For $n \in \{5, 7, 11, 13\}$ Wilson's theorem yields $(n - 1)! + 7 \equiv -1 + 7 \equiv 6 \pmod{n}$, which means that this expression cannot be a power of $n$. For $n = 9$ we obtain $0 \equiv 9^{k-1} \equiv 8! + 7 \equiv 1 \pmod{3}$, which again yields a contradiction. Finally, for $n \ge 15$ we note that $0 \equiv (n - 1)! + 7 \equiv n^{k-1} \pmod{7}$, which implies $7|n$. Since we obviously also have $7^2|(n - 1)!$ in this case, it follows that $0 \equiv n^{k-1} \equiv (n - 1)! + 7 \equiv 7 \pmod{7}$, which once more yields a contradiction.
The only case left to check is $k = 2$, but this yields the equation $(n - 1)! + 7 = n$, which cannot hold because of $(n - 1)! + 7 \ge (n - 1) + 7 > n$, and we see that the only two non-trivial solutions in this case are $(2, 4)$ and $(3, 3)$.

Now, let us consider the case $A = 2012$. We will show that there are no non-trivial solutions in this case.
For $n = 2$ the equation reduces to $2013 = 2^{k-1}$, which obviously has no integer solution. For $n > 2$, we have $2|(n - 1)!$, and therefore $n^{k-1} = (n - 1)! + 2012 \equiv 0 \pmod{2}$, which implies $n \equiv 0 \pmod{2}$. $n = 4$ yields $2018 = 4^{k-1}$ for the equation, which again has no integer solution. We therefore have only the case $n \ge 6$ left to consider.
If $k > 3$, the fact that $n$ is even yields $0 \equiv n^{k-1} \equiv (n - 1)! + 2012 \equiv 4 \pmod{8}$, which is a contradiction. For $k = 2$, the equation $(n - 1)! + 2012 = n$ cannot have a solution because $(n - 1)! + 2012 > n - 1 + 2012 > n$ certainly holds. The only case left is therefore $k = 3$, but this case does not yield a solution either, since $(n - 1)! + 2012 = n^2$ contradicts
$$(n - 1)! + 2012 > (n - 1)(n - 2)(n - 3) + 2012 = n^3 - 6n^2 - 18 + 2012 > n^2$$
(which is true because it is equivalent to $n^2(n - 7) + 12n + 1994 > 0$).
We see that there is no non-trivial solution for $A = 2012$, as claimed. $\square$