Hint: if $p,q>2$, we have $LHS=even, RHS=odd$. So we must have $p=2$, or $q=2$, casework.

Solution

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If both $p,q$ are odd than the $LHS$ is even and the $RHS$ is odd which we cannot have. So at least one of $p,q$ must be 2. Case 1:$p=2$ Then the equation becomes: $14q^2+2=q^3+344+1=q^3+345$ $14q^2-q^3=343$ $q^2(14-q)=343=49\cdot 7=7^2(14-7)$ So, $q=7$ Case 2:$q=2$ Then the equation becomes: $28p+p=8+43p^3+1=43p^3+9$ $A=29p=43p^3+9=B$ In this case since $p$ is odd A is odd but B is even which is a contradiction. So the only solution is $(p,q)=(2,7)$ (Same idea as #2 and #3)