Probably yes. The question is equivalent to: for any (nonzero) integer $n$, there exists an odd prime $p$ not dividing $n$ such that $p-1$ divides $4n$.

The divisors of $4$ are $1,2,4$; of these, $2,4$ are one less than odd primes $3,5$. So if $3\nmid n$ then choose $p=3$, and if $5\nmid n$ then choose $p=5$.

So we may suppose that $3,5$ both divide $n$. The divisors of $4\cdot3\cdot5=60$ that are one less than odd primes are $2, 4, 6, 10, 12, 30, 60$. So if $7\nmid n$ then choose $p=7$, if $11\nmid n$ then choose $p=11$, if $13\nmid n$ then choose $p=13$, if $31\nmid n$ then choose $p=31$, and if $61\nmid n$ then choose $p=61$.

So we may suppose that $3,5,7,11,13,31,61$ all divide $n$. The divisors of $4\cdot3\cdot5\cdot7\cdot11\cdot13\cdot31\cdot61$ that are one less than odd primes are $2, 4, 6, 10, 12, 22, 28, 30, 42, 52, 60, 66, 70, 78, 130, 156, 210, \
310, 330, 366, 372, 420, 462, 546, 660, 682, 732, 858, 910, 1092,
1302, 1612, 1708, 1830, 1860, 2002, 2310, 2730, 2860, 4026, 4092,
4270, 4620, 4758, 6006, 8052, 8580, 8866, 13420, 14322, 16926, 18910,
20020, 24180, 25620, 28182, 28210, 41602, 47580, 47740, 53196, 55510,
56730, 79422, 84630, 88660, 93940, 104676, 132370, 186186, 294996,
624030, 794220, 873642, 930930, 1221220, 1248060, 1474980, 1622478,
1831830, 1861860, 2912140, 3244956, 3663660, 11357346$; and so it continues....

All we need to prove that the answer to your question is "yes" is for this iterative procedure to contain infinitely many primes. Although that might be hard to prove, it would be incredible if the procedure halted.