We reduce to the following statement, which looks correct and should not be hard to prove.
Conjecture. Fix $n$ and $m$. Consider pairs of partitions $(\lambda,\mu)$ such that $\lambda$ has $m$ parts and $|\lambda|+|\nu|=m$. Let $A$ be the number of pairs for which $\max(\lambda)>\max(\mu)$ (where $\max(\emptyset)=-\infty$). Let $B$ be number of pairs for which $\lambda$ satisfies this condition with Durfee square, which we may rephrase as '$\lambda_i=i$ for some $i$, where $\lambda=(\lambda_1\geqslant \lambda_2\geqslant \dots)$'. Then $A=B$.
Let me give a bijection for $m=2$, maybe somebody can see how to generalize it. Take $\lambda=(\lambda_1\geqslant \lambda_2)$ and $\mu=(\mu_1\geqslant\dots)$ such that $\lambda_1>\mu_1$. There are three cases:
if $\lambda_2=1$ take partitions $(1,1)$ and $(\lambda_1-1\geqslant \mu_1\geqslant \dots)$;
if $\lambda_2=2$ take partitions $(\lambda_1\geqslant \lambda_2)$ and $(\mu_1\geqslant\dots)$;
if $\lambda_2\geq 3$ take partitions $(\lambda_2-1,2)$ and $(\lambda_1-1,\mu_1,\dots)$.
It is easy to see that this a bijection showing $A=B$ in this case $m=2$.
Now reduction.
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 their 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 $i$. Denote by $H(t,x)$ the part of double product $\prod_{n\geqslant 1} (1-tx^{n})(1-t^{-1}x^{n-1})$ with only non-negative powers of $t$. We have to prove that $$ \prod_{n\geqslant 1} (1-tx^n)\cdot H(t,x)=f(t,x)\cdot \prod (1-x^n)^{-1}. $$ Coefficient of $t^mx^n$ in RHS equals the number of pairs of partitions $(\lambda,\mu)$ for which $|\lambda|+|\mu|=n$, $\lambda$ has $m$ parts and is interesting and $\mu$ is arbitrary. By the conjecture it is the same as number of pairs of partitions $(\lambda,\mu)$ for which $|\lambda|+|\mu|=n$, $\lambda$ has $m$ parts and $\mu$ has a part which is not less then $\max(\lambda)$.
Let's obtain the same in the LHS. If $H(t,x)$ were not a part, but the whole product, then a lot would cancel when we multiple $\prod(1-tx^n)^{-1}$ and $\prod (1-tx^n)$. But 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 a 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 forbidden to replace $-tx^N$ to 1. This is what remains after removing all pairs of 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 choices from the second and from the third products, 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.
So, we choose a partition with maximal part $a_k$ (this is $\mu$) and another partition onto $m$ parts with maximal part at most $a_k$. This is $\lambda$. Also, exponent $m$ of $t$ is the number of parts in $\lambda$, and exponent of $x$ is $|\lambda|+|\nu|$.