Does $P \neq NP$ imply $NP \neq PSPACE$?

Is it true that
$\mathsf{P} \neq \mathsf{NP}$ implies $\mathsf{NP} \neq \mathsf{PSPACE}$?

I have here some problem that is in $\mathsf{PSPACE} \setminus \mathsf{NP}$ if $\mathsf{P} \neq \mathsf{NP}$. Since I cannot find a fault in my proof, I wanted to be sure, if the upper implication holds or not.

We know that $\mathrm{L}\subseteq\mathrm{NL}\subseteq\mathrm{P}\subseteq\mathrm{NP}\subseteq\mathrm{PSPACE}$ and at least one of those inclusions is strict, since $\mathrm{L}\subsetneq\mathrm{PSPACE}$ by the space hierarchy theorem. As far as I'm aware, any nonempty subset of the inclusions being strict would be consistent with our current state of knowledge so, in particular, it is not known that $\mathrm{P}\neq\mathrm{NP}$ implies $\mathrm{NP}\neq\mathrm{PSPACE}$.

StackExchange Computer Science Q#42108, answer score: 10