jmc

Squaring both sides, $$1+\sqrt{2y-3}={\left(\sqrt{1+\sqrt{2y-3}}\right)}^2={\left(\sqrt{6}\right)}^2=6.$$Hence, $\sqrt{2y-3}=5$. If we square this equation again, then $$2y-3={\left(\sqrt{2y-3}\right)}^2=5^2=25⟹y=\frac{25+3}{2}=\boxed{14}.$$