integer ordered pair,s $(x,y)$ in $1!+2!+3!+............+x! =y^3$ (1) Total no. of integer ordered pair,s $(x,y)$ in $1!+2!+3!+............+x! =y^2$
(2) Total no. of integer ordered pair,s $(x,y)$ in $1!+2!+3!+............+x! =y^3$
(3) Total no. of integer ordered pair,s $(x,y)$ in $1!+2!+3!+............+x! =y^4$
where $(x,y)\in \mathbb{N}.$
$\underline{\bf{My\;Try}}::$ for (1) one:: If $x\geq 4.$ Then $(1!+2!+3!+......+x!)$ conatin last digit is $ 3$ 
and we know that square of any integer does not have last digit is $3$.
So $x<4$, So we will check for $x = 1,2,3$.
If $x=1$, Then $y^2=1\Rightarrow y = 1$. So integer ordered pairs $(1,1)$.
If $x=2$, Then $y^2=3\Rightarrow y = \sqrt{3}$. So no integer ordered pairs.
If $x=3$, Then $y^2=9\Rightarrow y = 3$. So integer ordered pairs $(3,3)$.
So for (1) one integer ordered pair,s is $(x,y) = (1,1)\;\;,(3,3)$
But I did not understand how can i solve (2) and (3) one,
Help Required
Thanks

For Q1, you have already noticed that for $x = 4$, the LHS ends in $3$, and for $x > 4$, the terms that are added have last digit $0$.  As no square ends in $3$, we do not have any solution for $x \ge 4$.
For Q2, note that if $x \ge 9$, the LHS is divisible by $3$, so if this is the cube of an integer, that integer must also be divisible by $3$.  In fact the power must be divisible by $27$.  But $1! + 2! + ... + 8! \equiv 9 \pmod {27}$, hence there are no solutions for $x \ge 9$.
For Q3, you could notice that a fourth power is also a square and hence apply the answer to Q1, or extend the argument of Q2 (which in fact holds true for any larger power).
