Here's my method for looking at it:
For determining acidity, we need to look which one of $\ce{H2SO4}$ or $\ce{HNO3}$ form a more stable anion after the $H+$ is extracted.
Lets condsider $\ce{H2SO4}$ first-
• There are two sites to extract a proton.
• After a proton is removed, the $O-$ is in resonance with 2 oxygen atoms attached to the sulfur atom.
Now, in $\ce{HNO3}$ -
• There is one site to extract proton
• After removal of $H+$ , there are two resonance structures but the eletrons of the oxygen that lost a proton have delocalised only once and over a shorter region as compared to $\ce{H2SO4}$
Generally, a compound with more resonating structures is more stable due to more charge distribution. Amongst other factors like electronegativity, resonance usually dominates.
Fun fact:
$\ce{H2SO4}$ is stronger acid than $\ce{HNO3}$ , this is shown by following reaction:
$\ce{H2SO4} + \ce{HNO3} ---> \ce{NO2+} + \ce{HSO4-} + \ce{H2O}$
This reaction is used in the nitration of benzene to form the electrophile.