= Solution
Write an integral <binary quadratic form> as $[a,b,c]=ax^2+bxy+cy^2$. It is positive definite when $a>0$ and its <discriminant of a binary quadratic form>[discriminant] $D=b^2-4ac$ is negative. Two forms are <proper equivalence of binary quadratic forms>[properly equivalent] when one is obtained from the other by a change of variables in $\operatorname{SL}_2(\mathbb Z)$. A <reduced positive definite binary quadratic form> satisfies
$$
|b|\leq a\leq c,
$$
with $b\geq0$ when $|b|=a$ or $a=c$.
To prove reduction, choose in a proper equivalence class a form whose positive leading coefficient $a$ is minimal. Replacing $x$ by $x+ny$ changes $b$ by $2an$, so choose $n$ to arrange $|b|\leq a$. If the resulting $c<a$, the determinant-one substitution $(x,y)\mapsto(-y,x)$ would put $c$ in the leading position, contradicting minimality. Thus $a\leq c$. Sign changes on the boundary give the stated convention. Hence every positive definite form is properly equivalent to a reduced one.
A unimodular integral substitution is a bijection of $\mathbb Z^2$, so properly equivalent forms represent exactly the same integers.
Both given forms have discriminant $-23$. Reduction gives
$$
[6,5,2]\sim[2,1,3],
\qquad
[9,25,18]\sim[2,-1,3].
$$
The two reduced forms represent the same integers because
$$
[2,-1,3](x,y)=[2,1,3](x,-y),
$$
an integral change of variables of determinant $-1$. They are not properly equivalent: the uniqueness theorem for reduced positive definite forms says that each proper class has one reduced representative, and $[2,1,3]\ne[2,-1,3]$. Therefore the original forms represent the same integers but are not equivalent in the required proper sense.
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