Note that for every $n$, $n! + (n+1)! = n! + n! \cdot (n+1) = n! \cdot (n+2) \le (n+2)!$. Thus if $n! + (n+1)! = k! + 120$, then due to $120 = 5!$ we have $k! + 5! \le (n+2)!$. This inequality in turn implies $k < n+2$ and $5 < n+2$. Hence $5 \le n+1$, leading to $0 \le (n+1)! - 5! = k! - n!$. Consequently, $k \ge n$, i.e., the cases to be considered are $k = n$ and $k = n+1$. If $k = n$, then the initial equation leads to $(n+1)! = 120$, giving $n = 4$, $k = 4$. If $k = n+1$, then analogously $n = 5$, $k = 6$.

If $k < n$, then $n! > k!$. If, additionally, $n \ge 4$, then $(n+1)! \ge 120$, giving $n! + (n+1)! > k! + 120$. But if $n < 4$, then $n! + (n+1)! \le 30 < k! + 120$. The desired equality can hold in neither of the cases.
If $k > n+1$, then $k! - (n+1)!$ is positive and is also divisible by $(n+1)!$, hence $k! - (n+1)! \ge (n+1)!$. On the other hand, $n! - 120 < (n+1)!$, giving $k! - (n+1)! > n! - 120$. Thus, there is no solution in this case either.
Hence $n = k$ or $n+1 = k$, leading to two solutions $(n, k) = (4, 4), (5, 6)$.

The equation implies that $120$ is divisible by the minimum of $n!$ and $k!$. As $120 = 5!$, either $n \le 5$ or $k \le 5$. Consider both cases.
If $n \in \{1, 2, 3\}$, then $n! + (n+1)! - k! < 120$, whence the equation has no solution. If $n=4$, then $120 = 144 - k!$, whence $k! = 24$ and $k=4$. If $n=5$, then $120 = 840 - k!$, whence $k! = 720$ and $k=6$.
If $k \in \{1, 2, 3, 4, 5\}$, then the initial equation implies that $n! + (n+1)!$ lies between 121 and 240. This is possible only if $n=4$, since if $n=3$, then $n! + (n+1)! = 30$, and if $n=5$, then $n! + (n+1)! = 840$. If $n=4$, then $n! + (n+1)! = 144$, which corresponds to $k=4$.