Not an answer (yet?).
At first, let's count all the other partitions onto $m$ parts. These are partitions for which $k$'s largest part never equals $k$, $k=1,2,\dots$. We want to prove that the generating function equals $$ x^m\sum_{i=0}^m\frac{(-1)^{i}x^{i(i-1)/2}}{(x)_{m-i}} $$ (I have used that total number of partitions onto $m$ parts have generating function $x^m/(x)_m$, that is seen by duality.) Call such partitions interesting.
Multiply by $t^m$ and sum up by $m$. We get a double generating function $f(t,x)=\sum t^{{\rm parts}(\lambda)} x^{|\lambda|}$, summation is taken over all interesting partitions $\lambda$. We have to prove that $$ f(t,x)=\sum_{m\geqslant i\geqslant 0} (tx)^m\frac{(-1)^{i}x^{i(i-1)/2}}{(x)_{m-i}}= \left(\sum_{k\geqslant 0} \frac{(tx)^k}{(x)_k}\right)\left(\sum_{i\geqslant 0} (-1)^i(tx)^i x^{i(i-1)/2}\right). $$ As for the first multiple, it is a double generating function for $t^{\max(\lambda)} x^{|\lambda|}$ taken by all partitions $\lambda$, hence by duality it is the same thing as a double generating function for $t^{{\rm parts}(\lambda)} x^{|\lambda|}$, which is $\prod_{n\geqslant 1} (1-tx^n)^{-1}$. As for the second multiple, it is a part of Jacobi triple product $$ \prod_{n\geqslant 1} (1-tx^{n})(1-t^{-1}x^{n-1})(1-x^n)=\sum_{i=-\infty}^{\infty} (-1)^it^i x^{i(i+1)/2},$$ corresponding to non-negative $i$.
Now we have to do something. Well. What to do? If it were not a part of Jacobi triple product, but the whole JTP, then a lot would cancel when we multiple $\prod(1-tx^n)^{-1}$ and $\prod (1-tx^n)$. Now something still cancels! Namely, we look at our product $$ \prod_{n\geqslant 1} (1+tx^n+t^2x^{2n}+\dots) \prod_{n\geqslant 1} (1-tx^n) \prod_{n\geqslant 1} (1-t^{-1}x^{n-1}) $$ and see what we may take from each bracket. We are conditioned to take at least as many $t$'s from the second product than $t^{-1}$'s from the third. Consider partial involution on the set of our choices: denote by $N$ the maximal index $n$ for which we either take $-tx^{n}$ from the second product or take $t^ax^{na}$ for some $a\geqslant 1$ from the first product. We could take $1$ from the corresponding bracket in the second product and $t^{a+1}x^{n(a+1)}$ from the corresponding bracket in the first product instead. This is sign-changing involution on the set of choices, but the set of admissible choices is not quite invariant. Namely, if total number of $t$'s from the second product equals total number of $t^{-1}$'s in the third product, we are formidden to replace $-tx^N$ to 1. This is what remains after we cancel everything that cancels after removing all pairsof choices formed by involution. To be more precise, what we should choose are some $k$ positive integers $0<a_1<a_2<\dots <a_k$ (second product), some $k$ non-negative integers $0\leqslant b_1<b_2<\dots b_k$ (third product) and some partition with maximal part at most $a_k$ (first product).
As for two first choices, it is equivalent to the choice of a partition with maximal part $a_k$: for such a partition let $k$ be the size of Durfee square, $a_1,\dots,a_k$ correspond to the part of Young diagram on the one side of its diagonal and $b_1,\dots,b_k$ to the other side.
Oh. What we have reduced our claim to? We choose a partition with maximal part $a_k$ and another partition onto $m$ parts with maximal part at most $a_k$. It should be the same as choosing interesting partition with $m$ parts and another partition. Why? Or I have been mistaken somewhere above?