The original statement (number of partitions of $n$ where no part appears more than once equals to the number of partitions into parts not congruent to $\pm 1\pmod 6$) can be verified to be incorrect, such as in the cases $n=5$ and $n=7.$ The other statement in the comments (the number of partitions of $n$ in which no part appears exactly once equals the number of partitions into parts not congruent to $\pm 1\pmod 6$) holds and we will prove it here.
As with pretty much all proofs of such assertions about partitions, we will use generating functions. We wish to prove that $$\prod_{k=1}^{\infty}{\left(-x^k +\frac{1}{1-x^k}\right)}=\frac{\prod_{k=0}^{\infty }(1-x^{6k+1})(1-x^{6k+5})}{\prod_{k=1}^{\infty }(1-x^k)}.$$
Clearing the denominators, it is equivalent to prove that $$\prod_{k=1}^{\infty}{(1-x^k+x^{2k})}=\prod_{k=0}^{\infty }(1-x^{6k+1})(1-x^{6k+5}).$$
The secret ingredient to this proof is the fact that: The number of partitions of $n$ into parts that are all congruent to $\pm 1 \pmod{6}$ is equal to the number of partitions of $n$ into distinct parts that are all congruent to $\pm 1 \pmod{3}$. Here is a quick proof of this fact from p.4 of An Invitation to the Rogers-Ramanujan Identities by Andrew V. Sills: By difference of squares,
\begin{align*}
\prod_{k= 0}^{\infty}{(1+x^{3k+1})(1+x^{3k+2})} &= \frac{\prod_{k=0}^{\infty}{(1-x^{6k+2})(1-x^{6k+4})}}{\prod_{k=0}^{\infty}{(1-x^{3k+1})(1-x^{3k+2})}}\\
&= \frac{\prod_{k=0}^{\infty}{(1-x^{6k+2})(1-x^{6k+4})}}{\prod_{k=0}^{\infty}{(1-x^{6k+1})(1-x^{6k+4})(1-x^{6k+2})(1-x^{6k+5})}}\\
&= \frac {1}{\prod_{k=0}^{\infty }(1-x^{6k+1})(1-x^{6k+5})}.
\end{align*}
As a result, we know that
\begin{align*}
\prod_{k=0}^{\infty }(1-x^{6k+1})(1-x^{6k+5})&=\frac{1}{\prod_{k= 0}^{\infty}{(1+x^{3k+1})(1+x^{3k+2})}}\\
&=\frac{\prod_{k=1}^{\infty}{(1+x^{3k})}}{\prod_{k=1}^{\infty}{(1+x^k)}}.
\end{align*}
Thus, it suffices to prove that $$\prod_{k=1}^{\infty}{(1-x^k+x^{2k})}=\frac{\prod_{k=1}^{\infty}{(1+x^{3k})}}{\prod_{k=1}^{\infty}{(1+x^k)}}$$
or $$\prod_{k=1}^{\infty}{(1+x^k)(1-x^k+x^{2k})}=\prod_{k=1}^{\infty}{(1+x^{3k})}.$$
By sum of cubes, this is true because $$(1+x^k)(1-x^k+x^{2k})=1+x^{3k}.$$
Infinite products can be equivalent in such strange ways!
It sounds like you are asking for the number $p_k(n)$ of partitions of a number $n$ into exactly $k$ parts. This is a standard type of restricted partition problem and can be solved as follows. Given such a partition, "rotate the Ferrers diagram," which shows that $p_k(n)$ also counts the number of partitions of a number $n$ whose largest part has size exactly $k$. Removing this part produces a partition of $n-k$ whose largest part has size at most $k$, and using this we can show that for fixed $k$ this sequence has a very nice generating function given by
$$P_k(x) = \sum_{n \ge 0} p_k(n) x^n = \frac{x^k}{\prod_{i=1}^k (1 - x^i)}$$
which in principle allows you to write down a closed form for $p_k(n)$ (again, for fixed $k$; it gets more complicated the larger $k$ is), although I won't do so. As a random example, setting $k = 5$ gives
$$\begin{align} \sum_{n \ge 0} p_5(n) x^n &= \frac{x^5}{(1 - x)(1 - x^2)(1 - x^3)(1 - x^4)(1 - x^5)} \\ &= x^5 + x^6 + 2x^7 + 3x^8 + 5x^9 + 7x^{10} + 10x^{11} + 13x^{12} + 18x^{13} + \dots \end{align}$$
which shows, for example, that the number of partitions $p_5(13)$ of $13$ into exactly $5$ parts is $18$. $p_5(n-5)$, which also counts the number $p_{\le 5}(n)$ of partitions of $n$ into at most $5$ parts, turns out to be A001401 on the OEIS.
Among other things, this generating function can be used to deduce the asymptotic growth rate of $p_k(n)$ (for fixed $k$, as $n \to \infty$), which turns out to be
$$p_k(n) \sim \frac{n^{k-1}}{(k-1)! k!}.$$
Edit: Okay, mea culpa - I said I'd write down the closed form you get from the generating function but for large $k$ I am really not sure how to do it even in a messy way. The basic strategy is to compute the partial fraction decomposition of the generating function $P_k(x)$ above. This gets very messy in general; here's what you get for small values of $k$.
For $k = 1$ there is a unique partition of $n$ with $1$ part, namely $n = n$. This corresponds to the generating function
$$P_1(x) = \frac{x}{1 - x} = x + x^2 + x^3 + \dots $$
which gives $\boxed{ p_1(n) = 1 }$ if $n \ge 1$ and $0$ if $n = 0$. Pretty straightforward.
For $k = 2$ the partial fraction decomposition is
$$P_2(x) = \frac{x^2}{(1 - x)(1 - x^2)} = \frac{1}{2(1 - x)^2} - \frac{3}{4(1 - x)} + \frac{1}{4(1 + x)}$$
(this can be done by hand) which gives the already-surprisingly-complicated
$$\boxed{ p_2(n) = \frac{n}{2} + \frac{(-1)^n - 1}{4} }.$$
This can be simplified, and proven, with a little casework as follows. The partitions into two parts are $n = i + j$ where, if $i \ge j$, we have $1 \le j \le \lfloor \frac{n}{2} \rfloor$. So we get $\boxed{ p_2(n) = \left\lfloor \frac{n}{2} \right\rfloor }$ which is equivalent to the above after squinting a bit.
For $k = 3$ the partial fraction decomposition is (according to WolframAlpha)
$$P_3(x) = \frac{-x^2 + 20x - 7}{72(1 - x)^3} - \frac{8}{1 + x} + \frac{x + 2}{9(1 + x + x^2)}$$
which gives, using that $\frac{x + 2}{1 + x + x^2} = \frac{2 - x - x^2}{1 - x^3}$ and $-x^2 + 20x - 7 = - (1 - x)^2 - 18(1 - x) + 12$, and after some simplification,
$$\boxed{ p_3(n) = \frac{6n^2 - 7}{72} - \frac{(-1)^n}{8} + \frac{\omega_3(n)}{9} }$$
where $\omega_3(n)$ is a periodic function with period $3$ that repeats $2, -1, -1, 2, -1, -1, \dots $. Already it's not entirely obvious that this is an integer, and I can't claim not to have made any computational errors, but the result has to look something like this even if I messed something up. It only gets worse from here - $p_k(n)$ has period-$d$ components for every $1 \le d \le k$. You can check out the partial fraction decomposition for $k = 4$ for yourself here.
So there's an important sense in which it's hopeless to expect a closed form for large $k$, although as above the leading-order asymptotic in $n$ for fixed $k$ is easy to calculate and the next few terms might not be so bad either. The good news is that for many purposes it doesn't matter; the generating function is powerful enough to answer many questions you might have about partitions with restricted parts.
Best Answer
Let's define some $q$-series notation: $(a;q)_n = \prod_{i=0}^{n-1} (1 - aq^i)$ with shorthand notation $(q)_n = (q;q)_n$. E.g. the left-hand side of Euler's pentagonal theorem as stated in the question is $(x)_\infty$.
Now, $$P(n,k,m) = [x^n z^k] \prod_{i=1}^m \frac{1}{1 - x^i z} = [x^n z^k] \frac{\prod_{i=m}^\infty (1 - x^{i+1} z)}{\prod_{i=0}^\infty (1 - x^{i+1} z)} = [x^n z^k] \frac{(x^{m+1}z;x)_\infty}{(xz;x)_\infty}$$
There are a couple of theorems due to Euler: $$(-x;q)_\infty = \sum_{k=0}^\infty q^{k(k-1)/2}(q)_k^{-1} x^k \\ (-x;q)_\infty^{-1} = \sum_{k=0}^\infty (-1)^k (q)_k^{-1} x^k$$
Then $$(x^{m+1}z;x)_\infty(xz;x)_\infty^{-1} = \left(\sum_{k=0}^\infty x^{k(k-1)/2}(x)_k^{-1} (-x^{m+1}z)^k\right) \left(\sum_{k=0}^\infty (-1)^k (x)_k^{-1} (-xz)^k\right) \\ = \left(\sum_{j=0}^\infty (-1)^j x^{j(j+2m+1)/2}(x)_j^{-1} z^j\right) \left(\sum_{k=0}^\infty (x)_k^{-1} x^k z^k\right) \\ = \sum_{j=0}^\infty \sum_{k=0}^\infty \frac{(-1)^j x^{j(j+2m+1)/2+k} z^{j+k}}{(x)_j (x)_k}$$
In particular, $$P(n,k,m) = [x^n] \sum_{j=0}^\infty \frac{(-1)^j x^{j(j+2m-1)/2+k}}{(x)_j (x)_{k-j}}$$
Note that if $P'(n,m)$ counts the partitions of $n$ into parts of at most $m$ then $$P'(n,m) = [x^n] \prod_{i=1}^m \frac{1}{1-x^i} = [x^n] (x)_m^{-1}$$ so $$P(n,k,m) = \sum_{a=0}^\infty \sum_{b=0}^\infty \sum_{j=0}^\infty (-1)^j P'(a,j) P'(b,k-j) [n-k-a-b=j(j+2m-1)/2]\\ = \sum_{a=0}^\infty \sum_{b=0}^\infty \sum_{j \in \frac{1-2m\pm\sqrt{4m^2-4m+1+8(n-k-a-b)}}{2}} (-1)^j P'(a,j) P'(b,k-j)$$
It's not quite as elegant as the pentagonal theorem...