Is the number of primes equal to the number of integers? I understand countable infinities and that any subset of the integers should be an equivalent countable infinity, but the primes seem so odd (sorry, weird!), and I haven't found anything directly addressing this, so I thought the answer might include something interesting. Or is it just that we know there is no largest prime, and that's the end of it? It seems to me that the idea that there is also no largest gap between primes might affect this.
 A: We know that there is no largest prime. We know that $p_1=2$ is the first prime, and if we have the $n$th prime $p_n$, then because there are primes $>p_n$ (and these primes are natural numbers), there is a smallest prime among the primes $>p_n$, which we can define as $(n+1)$st prime $p_{n+1}$. This way, we do indeed efficiently get a never-ending enumeration of the primes - that is, a bijection between the set of natural numbers and the set of primes. That is what we mean by there are "equally many" primes as there are naturals.
A: "I understand countable infinities and that any (infinite [ed]) subset of the integers should be an equivalent countable infinity"
ANd that's all there is to it.  The primes are a subset of the natural numbers.  There are infinitely many primes.  So there cardinality of the primes is the same as that of the natural numbers.
If you want to get the bijection we can do so by letting $p_1=2$ the first prime and $p_2 = 3$ the second prime and so on.  $p_{k+1}$ is the next prime after $p_k$.  Then $n\mapsto p_n$ is the bijection between $\mathbb N$ and $\{$primes$\}$.
That's really all there is to it.  

t seems to me that the idea that there is also no largest gap between primes might affect this.

Not really.  There is a largest gap between odds, but not between primes (or, powers of $17$ just to give an example).  
Countable is the "smallest" infinite cardinality.  As such every infinite subset has the same cardinality and it doesn't really matter how densely or sparsely we pack them into another subset.  (Which if you think about it why would how the order in the integers matter any more than how you order them in, say the rationals, and we can reorder them without changing the cardinality).
It's good to wonder about these issues but sometimes they just boil to ... nothing much to say.
A: Like hardmath has mentioned, one could define a bijective map $f:\mathbb{Z}\rightarrow \text{primes}$ simply by mapping integer to $f(0)=p_0=2,f(1)=p_1=3...$ very much because integer was a well ordered set. 
Thus, in such sense, you could claim that they were equal because the number of integer and number of primes was of the same order. However, be aware that since the order of the set was not finite, the explanation had to be done with set theory. 
