jmc

We have $$\begin{array}{rl}{b}^2& =121_c\\ & =c^2+2c+1\\ & =(c+1{)}^2,\end{array}$$so $b=c+1$ (we don't consider the "solution" $b=-(c+1)$ since $b$ and $c$ are both required to be positive).

We also have $$\begin{array}{rl}{c}^2& =71_b\\ & =7b+1.\end{array}$$We also know that $c=b-1$, so $c^2=(b-1{)}^2=b^2-2b+1$. Thus $$b^2-2b+1=7b+1.$$Adding $2b-1$ to both sides, we have $$b^2=9b.$$The only positive solution is $b=9$, which gives $c=8$ and thus $b+c=\boxed{17}$.