Let $p$ be a prime, and let $q$ be a prime other than $p$. Then the congruence $\lambda p+1\equiv 0\pmod{q}$ always has a solution. So there are no primes $p$ with the property you are asking for.
Remark: If you are not familiar with properties of congruences, we prove the result without referring to such properties.
Look at the remainders when the numbers $0\cdot p+1$, $1\cdot p+1$, $2\cdot p+1$, and so on up to $(q-1)\cdot p+1$ are divided by $q$. These remainders are all different. For if $i\cdot p+1$ has the same remainder as $j\cdot p+1$, then $q$ divides $(i\cdot p+1)-(j\cdot p+1)$, so $q$ divides $(i-j)\cdot p$. But since $q\ne p$, it follows that $q$ divides $i-j$. Since $0\le i,j\le q-1$, this forces $i=j$.
Note that we did not need here that $p$ and $q$ are prime. It is enough to assume that $p$ and $q$ have no common divisor greater than $1$ (in other words, that $p$ and $q$ are relatively prime.)
Since the remainders of $0\cdot p+1$, $1\cdot p+1$, $2\cdot p+1$, and so on up to $(q-1)\cdot p+1$ are all different, they must be $0$, $1$, $2$, and so on up to $q-1$, in some order. In particular, there is an $i$ such that the remainder when you divide $i\cdot p+1$ by $q$ is $0$.