Your equation can be rewritten as
$$ (x+y-d)^2+(2d-1)(x-y)^2=d(2f+d). $$
This shows that the number of integral solutions $(x,y)$ is much the same as the number of divisors of the right hand side in the ring of integers of $\mathbb{Q}(\sqrt{1-2d})$. Of course there are issues of parity to consider, but these are not too relevant. For general $d$ it is quite subtle to give an explicit formula for the number of divisors , because the class group is nontrivial (I recommend the book Cox: Primes of the form $x^2+ny^2$). At any rate, the number of solutions can be as large as $\exp(c\log f/\log\log f)$ for a constant $c=c(d)>0$, and it is easy to construct concrete examples when the number of solutions is much larger than $10$. I will give examples in the next paragraph (I plan to add another paragraph later, for slightly larger $d$).

For $d=2$ we are working in the ring of integers of $\mathbb{Q}(\sqrt{-3})$ which is a unique factorization domain. It is known that in this ring any rational prime $\equiv 1\pmod{3}$ splits into a product of two primes. This means that if $2f+d=2f+2$ has many such prime factors, the number of divisors will be large. For example, for $f=7*13*19*31-1=53598$, SAGE gives me $96$ solutions starting with $(-266,-144)$ and ending with $(268,146)$. For $f=7*13*19*31*37-1=1983162$, SAGE gives me $192$ solutions starting with $(-1625,-828)$ and ending with $(1627,830)$.

For $d=10$ we are working in the ring of integers of $\mathbb{Q}(\sqrt{-19})$ which is a unique factorization domain. It is known that in this ring any rational prime $\equiv 1, 4, 5, 6, 7, 9, 11, 16, 17\pmod{19}$ splits into a product of two primes. This means that if $2f+d=2f+10$ has many such prime factors, the number of divisors will be large. For example, for $f=5*7*11*17*23-5=150530$, SAGE gives me $96$ solutions starting with $(-885,-802)$ and ending with $(895,812)$. For $f=5*7*11*17*23*43-5=6473000$, SAGE gives me $192$ solutions starting with $(-5831,-5290)$ and ending with $(5841,5300)$.

**Added.** It turns out that the original question concerned the special case $f=0$. I discussed that case in a second answer that was incorrect, unfortunately. Here I collect what could be salvaged from that answer as it might be useful for later investigations. The equation can be rewritten as
$$ (2d-1)(x-y)^2=(x+y)(2d-x-y), $$
where all factors can be seen to be nonnegative integers. The case $x=y$ yields the lattice points $(0,0)$ and $(d,d)$. In the case $x\neq y$ we have a factorization into positive integers
$$2d-1=d_1d_2,\quad (x-y)^2=b_1b_2,\quad x+y=d_1b_1,\quad 2d-x-y=d_2b_2,$$
and the case when $d_1$ or $d_2$ equals $1$ yields the other four listed lattice points $(0,1)$, $(1,0)$, $(d-1,d)$, $(d,d-1)$. These observations certainly show (without any algebraic number theory) that the number of solutions is $O_\epsilon(d^\epsilon)$, and a good approach seems to consider the system
$$ d_1d_2=2d-1,\quad b_1b_2=\square, \quad d_1b_1+d_2b_2=2d. $$
It is worthwhile to note that $d_1$ and $d_2$ are coprime here, which implies that the six listed lattice points are the only ones when $2d-1$ is a prime power.