= Solution
Write an integral <binary quadratic form> as $[a,b,c]$. For <proper equivalence of binary quadratic forms>, the allowed changes of variables have <determinant> one. Here is the needed reduction rather than an appeal to class-number tables. A shear $(x,y)\mapsto(x+ny,y)$ changes $b$ to $b+2an$ without changing $a$ or the <discriminant of a binary quadratic form>. Choose $n$ so $-a<b\le a$. If the resulting $c<a$, the determinant-one change $(x,y)\mapsto(-y,x)$ replaces $[a,b,c]$ by $[c,-b,a]$, strictly decreasing the positive integer leading coefficient. Repeating must terminate with $|b|\le a\le c$. If $a=c$ and $b<0$, the same rotation makes $b$ positive; the choice $-a<b\le a$ already fixes the other boundary sign. This is a <reduced positive definite binary quadratic form>.
Since $4ac-b^2=67$, the reduction inequalities give $67\ge3a^2$, so $a\in\{1,2,3,4\}$. The integer $b$ is odd. For $a=1$, only $b=1$ survives the boundary convention and $c=17$. For $a=2$, $b=\pm1$ gives $c=68/8$, not an integer. For $a=3$, $b=\pm1,\pm3$ gives $68/12$ or $76/12$; for $a=4$, $b=\pm1,\pm3$ gives $68/16$ or $76/16$. None is integral. Therefore \b[every form is properly equivalent to $x^2+xy+17y^2$], establishing <class number one for discriminant minus sixty-seven>.
For infinitude, its determinant-one shears produce
$$
\boxed{f_n(x,y)=x^2+(2n+1)xy+(n^2+n+17)y^2,\qquad n\in\mathbb Z.}
$$
Each has discriminant $-67$ and is positive definite because it is a change of variables of the displayed positive form. Their middle coefficients are distinct, so they are infinitely many distinct elements of the set.
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