Fedja is absolutely right: this has been proven, for sufficiently large $n$, by Drmota, Mauduit and Rivat.
Although it looks at first sight as though this question is as hopeless as any other famous open problem on primes, it is easy to explain why this is not the case. Of the numbers between $1$ and $N := 2^{2n}$, the proportion whose digit sum is precisely $n$ is a constant over $\sqrt{\log N}$. These numbers are therefore quite "dense", and there is a technique in prime number theory called the method of bilinear sums (or the method of Type I/II sums) which in principle allow one to seriously think about finding primes in such a set. This is what Drmota, Mauduit and Rivat do.
I do not believe that their method has currently been pushed as far as (for example) showing that there are infinitely many primes with no 0 when written in base 1000000.
Let me also remark that they depend in a really weird way on some specific properties of these digit representation functions, in particular concerning the sum of the absolute values of their Fourier coefficients, which is surprisingly small. That is, it is not the case that they treat these Hamming sets as though they were "typical" sets of density $1/\sqrt{\log N}$.
I think one might also mention a celebrated paper of Friedlander and Iwaniec, http://arxiv.org/abs/math/9811185. In this work they prove that there are infinitely many primes of the form $x^2 + y^4$. This sequence has density just $c/N^{1/4}$ in the numbers up to $N$, so the analysis necessary to make the bilinear forms method work is really tough. Slightly later, Heath-Brown adapted their ideas to handle $x^3 + 2y^3$. Maybe that's in some sense the sparsest explicit sequence in which infinitely many primes are known (except of course for silly sequences like $s_n$ equals the first prime bigger than $2^{2^n}$).
Finally, let me add the following: proving that, for some fixed $n$, there are infinitely many primes which are the sum of $n$ powers of two - this is almost certainly an open problem of the same kind of difficulty as Mersenne primes and so on.
Let me first comment on your examples:
The record is acually $t=n^{0.525}$, due to Baker, Harman and Pintz.
A slightly thinner sequence of this type is $n=a^3+2b^3$, investigated by
Heath-Brown, and also cubic polynomials in two variables,
by Heath-Brown and Moroz,
eprints.maths.ox.ac.uk/00000163/01/morozcub2.pdf
Other polynomials in two or more variables, like those that you suggest,
are certainly not known by today's methods.
- Here is a sketch to obtain primes with a density of
0.7375 one bits (compare also this paper by Eric Naslund
https://arxiv.org/abs/1211.2455
and his answer on mathoverflow, to a question by Gil Kalai
Primes with more ones than zeroes in their Binary expansion
their-binary-expansion/97345#97345
By Baker-Harman-Pintz one can fix the first 0.475n bits as one.
With $x=2^n$
There are $\gg x/\log x$ many primes in the interval [2^n-(2^n)^0.525, 2^n].
For the remaining 0.525n bits about 50% must be 0 and 50% ones.
If the proportion of one bits was smaller, then estimating the tail
(sum of binomial coefficients) would show the number of primes is too small.
So altogether the asymptotic density of one bits can be as large as
$(0.475+1/2\, 0.525)n=0.7375n$.
Here is a paper, where a similar method was used to find quadratic non-squares
or even primitive roots, with only few one-bits (less bits than the least non-square or least primitive root might need).
Now an example for a sequence with quite small
counting function (Problem 1).
Let us first look at large and sparse, but finite sets.
Let $F_n=2^{2^n}+1$ be the $n$-th Fermat number.
Let $D(F_n)$ be the set of its divisors.
Observe that the number of divisors is (by Wigert's bound)
$d(F_n)=O(exp (c \frac{\log F_n}{\log \log F_n})=O(\exp(c' 2^n/n)$.
Let $S(k)=\cup_{n=1}^k D(F_n)\subset [1,2^{2^k}+1]$.
Observe that $|S(k)|= O(k \exp(c' 2^k /k)< (F_k)^{0.499}$.
The set of divisors of the Fermat numbers contains
also the prime factors (for each $F_n$ at least one new prime number,
as the Fermat numbers are coprime).
Well, to extend this to an infinite set we need to take care of the fact
that with $F_{n+1}$ one possibly also adds smaller elements, so that
the counting becomes a bit more tricky.
But these added elements are not very small,
as one can show that the prime factors of $F_n$ are
at least of size $2^n$.
So maybe one can take the union over a thin subset of
the Fermat numbers, such as $\cup_{n=1}^k D(F_{2^n})$ or similar,
Or one only takes the second smallest element $s_n$ of $D(F_n)$,
(which must be a prime!).
$S= \cup_{n=1}^{\infty} s_n$.
Note again, that the $s_n$ are distinct and are (at least)
exponentially increasing, $s_n>2^n$, by properties of the Fermat numbers.
In case anybody thinks this is an artificial sequence, equivalent to writing
down some prime numbers:
well, the prime divisors of sequences such as
$s_n=n^2+1$ or $s_n=\lfloor 2^n/n\rfloor$
might not work, as these might be too dense.
Best Answer
Almost all $n$ are insipid. In fact, the number of non-insipid numbers at most $n$ grows like $2n/\log n$. See the paper
Cameron, Peter J.; Neumann, Peter M.; Teague, David N. On the degrees of primitive permutation groups. Math. Z. 180 (1982), 141–149. doi.org/10.1007/BF01318900