Hint $\rm\, 1575 = 15\!\cdot\! 100\! +\! 75 = 25\,(15\!\cdot\! 4\! +\! 3) = 5^2\cdot 3^2\cdot\color{#C00} 7\:$ lacks only one $\,\color{#C00}{??}\,$ to become square.
Remark $\ $ Suppose, instead, that $\rm\:n\:$ is not $\,1575\,$ but is a bigger integer that is difficult to factor completely, but we are given that it is not square. First, we can rule out $\rm\:k = 9\,$ and $\,25,\,$ since multiplying $\rm\:n\:$ by a square does not change the squareness of $\rm\:n.\:$ Since $\,63 = 7\cdot 3^2,\:$ we infer that if $\rm\:63\,n = 3^2\!\cdot\! 7\,n\:$ is a square then so too is the smaller $\rm\:7n.\:$ This leaves only the possibilities $\rm\: k = 7\:$ or $\rm\:15 = 3\!\cdot\! 5.\:$ To determine which $\rm\:kn\:$ is square, we need only determine the parity of the power of any one of the primes $\,3,5,7\,$ in the factorization of $\rm\:n.\:$ For example, in your case, once we have determined that $\rm\:n = 5^2\,j,\ 5\nmid j,\:$ then we know $\rm\:k\:$ must have have an even power of $5$, which excludes $\rm\:k=3\!\cdot\!5,\:$ leaving $\rm\:k = 7\:$ as the only possible solution.
It involves solving the equation $x^2-48y^2=1$ in integers $x,y$ since it has a smallest solution $(7,1)$ we get infinitely many solutions $(x_n,y_n)$ from the relations $(x_n+y_n\sqrt{48})=(7+\sqrt{8})^n$.And if you don't know how I reached this relation then visit the following link:
http://en.wikipedia.org/wiki/Pell%27s_equation
Best Answer
We’ll first want to convert this into a Pell equation. This is an equation of the form $x^2-cy^2=1$. To do this, we do some algebraic manipulation (which essentially amounts to completing the square):$$2n^2+3n+1=6k^2\Leftrightarrow$$ $$16n^2+24n+8=48k^2\Leftrightarrow$$ $$(4n+3)^2-3(4k)^2=1.$$ Letting $x=4n+3$, $y=4k$, $c=3$, our equation takes exactly the form we needed.
Now, Pell equations have been thoroughly studied, and standard methods exist to enumerate its solutions, in order. We just need to enumerate the solutions of this one until $n$ and $k$ take integer values. By doing this, the first solution we arrive at is $(x,y)=(7,4)$, but the condition $n>1$ rejects it. The next solution we arrive at is $(x,y)=(1351,780)$, which gives $(n,k)=(337,195)$, as our smallest solution.