$8(1!+2!+...+k!)+1=(2n+1)^2$ If $k>=10$ With $mod5$ we have $5|LHS$ so $25|LHS$ contradiction We left with the cases $k<10$

We claim the solutions are $(k,n)=(1, 1), (2, 2), (5, 17)$. If $k\ge 7$, then $LHS \equiv 5 \pmod 7$. However, we see $\frac{n(n+1)}{2}\equiv 0, 1, 3, 6 \pmod 7$. Thus, $k\le 6$. Hence, we consider these to obtain the solutions.

We take mod 7. Claim 1: If $k \ge 7$, then, $1! + 2! + ... + k! \equiv 5\pmod{7}.$ Proof: $1! + 2! + ... + k! \equiv 1!+2!+3!+4!+5!+6! \equiv 5\pmod{7}.$ Also, we can see that, $\frac{n(n+1)}{2}\equiv 0, 1, 3, 6 \pmod 7$, is always true, for $k \ge 7$, now, we have contradiction, hence, $k \le 7.$ Now, if $k=1$, then $n=1$, if $k=2$, then $n=2$, if $k=3,4$, no solutions, if $k=5,$ then $n=17$, so $(k,n)=\boxed{(1,1), (2,2), (5,17)}.$

Let's look at mod 7. $ 1!+2!+ \cdot\cdot\cdot k! =5 (mod7) $ $\frac{n(n+1)}{2} =0,1,3,6 (mod7) $ contraction! So, k <7 it only works in $ (k,n)=(1,1)(2,2)(5,17) $

$\boxed{ANSWER:(k,n)=(1,1)(2,2)(5,17)}$ claim:$k<7$ proof: if $k>7\implies LHS\equiv 5\pmod{7}$ and $RHS\equiv 0,1,3,6\pmod{7}$ which is contradiction $k<7\implies k=1,2,5$