$1!=1^2$ giving the solution $(1,1)$ $1!+2!=3$ which is not a square. $1!+2!+3!=3^2$ giving the solution $(3,3)$ $1!+2!+3!+4!=33$ which is not a square. Since $10|k!$ for all $k \geq 5$, $1!+2!+ \dots n! \equiv 3 \mod 10$ for all $n \geq 4$. But there are no perfect squares congruent to 3 modulo 10: hence the only solutions are those already found.

We have that any square is congruent to $ 0,1,4,5,6$ or $ 9$ modulo $ 10$. But in the expression given for $ n$ graeter than $ 3$ we get the expression to be congruent to $ 3$ modulo $ 10$. So, the possible values for $ n$ and $ m$ are $ 1$ and $ 3$ only!

How is that a different solution? That's exactly the same, or am I missing something?

Sorry, I did not see the prevoius post. Anyway i was thinking can we replace $ m^{2}$ with $ m^{k}$ where $ k$ is greater than $ 2$ ?

For k>2, we can look at the sum modulo 27 for n=8: $ 1+2+6-3+12-9+18+9\equiv9\pmod{27}$ Since for all $ n>8$ $ 27|n!$, we have that $ S_n\equiv9\pmod{27}\forall n>7$. Only for $ n=7$ $ 27|S_7$, but $ S_7=5913=3^4\cdot73$, so there are no solution for $ k>2$