Show it is impossible to find a $+ve$ integer such that sum of its sq. and its cube is an integral multiple of the sq. of the next highest integer. Based on the problem statement, we can form the equation as :-

$x^2+x^3 = k(x+1)^2$

where $x$ and $x+1$ are positive integers and $k$ will be the integral multiple. 
One doubt that I have here is as per the statement " $k$ will be the integral multiple" , does this mean that $k$ will be a Whole number i.e. $k$ can be a positive integer, negative integer or $0$ ?
Now

$x^2+x^3 = k(x+1)^2$  $\Rightarrow x^2(x+1) = k(x+1)^2$ 
$\Rightarrow (x+1) [x^2-k(x+1)] = 0$

One solution will be $x =-1$ which is not a valid solution in this case.
Another solution will come from $x^2-k(x+1) =0$ which gives $x$ as :- 

$x = \frac{k}{2} \pm \sqrt{\frac{k(k+4)}{4}}$

How can we proceed from here in order to prove the statement? One way I was adopting was to put different integral values of $k$ and checking if we get any $x$ as a positive integer but this does not seem to be a reliable way to prove this. I am stuck at this part. Please help !
Thanks in advance !

An easier approach might be to say
$$k = \frac{x^2+x^3}{(x+1)^2} = \frac{x^2(1+x)}{(x+1)^2}= \frac{x^2}{x+1}= \frac{x^2-1}{x+1}+\frac{1}{x+1} =x-1+\frac{1}{x+1}$$
and, for positive integer $x$, that is strictly between $x-1$ and $x$, so not an integer
